Differentiation of Polynomials

P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3

CHAPTER 19 Differentiation of Polynomials

Objectives

To understand the concept of limit. To understand the definition of differentiation. To understand and use the notation for the derivative of a polynomial function. To be able to find the gradient of a curve of a polynomial function by calculating its derivative. To apply the rules for differentiating polynomials to solving problems.

In the previous chapter the rate of change of one quantity with respect to a second quantity has been considered. In this chapter a technique will be developed for calculating the rate of change for polynomial functions. To illustrate this an introductory example will be considered. On planet X,anobject falls a distance of y metres in t seconds, where y = 0.8t2. (Note: On Earth the commonly used model is y = 4.9t2.) Can a general expression for the speed of such an object after t seconds be found? In the previous chapter it was y found that the gradient of the curve y = 0.8t 2 at a given point P can be approximated by ﬁnding the gradient of a chord PQ, where Q is a point on the curve as close Q as possible to P. The gradient of chord PQ approximates SAMPLEP the speed of the object at P. The ‘shorter’ this chord is made the better 0 5 (5 + h) t the approximation.

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Chapter 19 — Differentiation of Polynomials 537

Let P be the point on the curve where t = 5. Let Q be the point on the curve corresponding to h seconds after t = 5, i.e. Q is the point on the curve where t = 5 + h. 0.8 (5 + h)2 − 52 The gradient of chord PQ = (5 + h) − 5 0.8 (5 + h)2 − 52 = h = 0.8(10 + h) The table gives the gradient for different values of h. h Gradient of PQ Use your calculator to check these. If values of h of smaller and smaller magnitude are 0.7 8.56 taken, it is found that the gradient of chord PQ gets 0.6 8.48 = closer and closer to 8. At the point where t 5 the 0.5 8.40 gradient is 8. Thus the speed of the object at the 0.4 8.32 = moment t 5is8metres per second. 0.3 8.24 = The speed of the object at the moment t 5isthe 0.2 8.16 limiting value of the gradients of PQ as Q approaches P. 0.1 8.08

A formula for the speed of the object at any time t is required. Let P be the point with coordinates (t, 0.8t2)onthecurve and Q be the point with coordinates (t + h, 0.8(t + h)2). 0.8 (t + h)2 − t2 The gradient of chord PQ = (t + h) − t = 0.8(2t + h) From this an expression for the speed can be found. Consider the limit as h approaches 0, that is, the value of 0.8(2t + h)ash becomes arbitrarily small. The speed at time t is 1.6t metres per second. (The gradient of the curve at the point corresponding to time t is 1.6t.) This technique can be used to investigate the gradient of similar functions. 19.1 The gradient of a curve at a point, and the gradient function Consider the function f : R →R, f (x) = x2. y The gradient of the chord PQ in the f (x) = x2 adjacent graph (a + h, (a + h)2) Q + 2 − 2 = (a h) a a + h − a a2 + 2ah + h2 − a2 SAMPLE= 2 a + h − a P(a, a ) x = 2a + h 0 and the gradient at P can be seen to be 2a. The limit of (2a + h)ash approaches 0 is 2a. It can be seen that there is nothing special about a.Soifx is a real number a similar formula holds. Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3

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The gradient of the function y = x2 at any point x is equal to 2x. It is said that 2x is the derivative of x2 with respect to x or, more brieﬂy, the derivative of x2 is 2x. This itself is clearly the rule for a function of x and we refer to this function as the gradient (or derived) function. The straight line that passes through P and has gradient 2a is called the tangent to the curve at P. From the discussion at the beginning of the chapter it can be seen that the derivative of 0.8t2 is 1.6t.

Example 1

By ﬁrst considering the gradient of the chord QP, ﬁnd the gradient of y = x2 − 2x at the point Q with coordinates (3, 3).

Solution y Consider chord PQ: y = x2 – 2x Gradient of PQ + 2 − + − (3 + h, (3 + h)2 – 2(3 + h)) = (3 h) 2(3 h) 3 P 3 + h − 3 + + 2 − − − = 9 6h h 6 2h 3 3 + h − 3 Q(3, 3) + 2 = 4h h h x = 4 + h 0 2

Consider h as it approaches zero. The gradient at the point (3, 3) is 4.

Example 2

Find the gradient of chord PQ and hence the derivative of x2 + x.

y = x2 + x

P(x + h, (x + h)2 + (x + h))

Q(x, x2 + x) SAMPLEx –1 0

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Chapter 19 — Differentiation of Polynomials 539

Solution (x + h)2 + (x + h) − (x2 + x) The gradient of chord PQ = x + h − x 2 + + 2 + + − 2 − = x 2xh h x h x x h + 2 + = 2xh h h h = 2x + h + 1

From this it is seen that the derivative of x2 + x is 2x + 1.

The notation for limit of 2x + h + 1ash approaches 0 is lim 2x + h + 1. h→0 The derivative of a function with rule f (x)may be found by first finding an expression for the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) and then finding the limit of this expression as h approaches 0.

Example 3

By ﬁrst considering the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) for the curve f (x) = x3, ﬁnd the derivative of x3.

Solution f (x) = x3 f (x + h) = (x + h)3 f (x + h) − f (x) The gradient of chord PQ = (x + h) − x + 3 − 3 = (x h) x (x + h) − x (x + h)3 − x3 The derivative of f (x) = lim h→0 (x + h) − x x3 + 3x2h + 3h2x + h3 − x3 = lim h→0 h 3x2h + 3h2x + h3 = lim h→0 h = lim 3x2 + 3hx + h2 h→0 SAMPLE= 3x2 We have found that the derivative of x3 is 3x2. The following example provides practice in determining limits.

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Example 4

Find: 3x2h + 2h2 a lim 22x2 + 20xh + h b lim h→0 h→0 h c lim 3x d lim 4 h→0 h→0

Solution 3x2h + 2h2 a lim 22x2 + 20xh + h = 22x2 b lim = lim 3x2 + 2h h→0 h→0 h h→0 = 3x2 c lim 3x = 3x d lim 4 = 4 h→0 h→0

Using the TI-Nspire b 11 Use Deﬁne√ ( ) with the function f (x) = x and then complete as shown.

Using the Casio ClassPad For Example 4b, enter and highlight the expression (3x2h + h2)/h then from the keyboard select 2D—Calc— and enter h and SAMPLE0inthe spaces provided, as shown.

Chapter 19 — Differentiation of Polynomials 541

Exercise 19A

1 A space vehicle moves so that the distance travelled over its ﬁrst minute of motion is given by y = 4t4,where y is the distance travelled in metres and t the time in seconds. By ﬁnding the gradient of the chord between the points where t = 4 and t = 5, estimate the speed of the space vehicle when t = 5.

2 A population of insects grows so that the population, P,attime t (days) is given by P = 1000 + t2 + t,where t > 0. By ﬁnding the gradient of the chord between the points where t = 3 and t = 3 + h, ﬁnd an estimate for the rate of growth of the insect population at time t = 3.

Example 4 3 Find: 2x2h3 + xh2 + h 3x2h − 2xh2 + h a lim b lim c lim 20 − 10h h→0 h h→0 h h→0 30hx2 + 2h2 + h d lim e lim 5 h→0 h h→0 Example 3 4 Find: (x + h)2 + 2(x + h) − (x2 + 2x) a lim i.e. the derivative of y = x2 + 2x h→0 h (5 + h)2 + 3(5 + h) − 40 b lim i.e. the gradient of y = x2 + 3x,where x = 5 h→0 h (x + h)3 + 2(x + h)2 − (x3 + 2x2) c lim i.e. the derivative of y = x3 + 2x2 h→0 h Example 1 5 Foracurve with equation y = 3x2 − x: a find the gradient of chord PQ,where P is the point (1, 2) and Q is the point ((1 + h), 3(1 + h)2 − (1 + h)) b find the gradient of PQ when h = 0.1 c find the gradient of the curve at P. 2 6 Foracurve with equation y = : x a find the gradient of chord AB where A is the point (2, 1) and B is the point 2 (2 + h), 2 + h b find the gradient of AB when h = 0.1 c find the gradient of the curve at A. SAMPLE7 Foracurve with equation y = x2 + 2x − 3: a find the gradient of chord PQ,where P is the point (2, 5) and Q is the point ((2 + h), (2 + h)2 + 2(2 + h) −3) b find the gradient of PQ when h = 0.1 c find the gradient of the curve at P.

542 Essential Mathematical Methods1&2CAS 19.2 The derived function In Section 19.1 we saw how the gradient function of a function with rule f (x) could be derived by considering the gradient of a chord PQ on the curve of y = f (x). Consider the graph y = f (x)ofthe function f : R → R. f (x + h) − f (x) y The gradient of the chord PQ = x + h − x y = f (x) + − = f (x h) f (x) h Q Therefore the gradient of the graph at P is (x + h, f (x + h)) P (x, f (x)) given by x f (x + h) − f (x) 0 lim h→0 h

The reader is referred to Section 19.4 for a further discussion of limits.

The gradient or derived function is denoted by f , f (x + h) − f (x) where f : R → R and f (x) = lim h→0 h

In this chapter only polynomial functions are considered. For a polynomial function the derived function always exists and is deﬁned for every number in the domain of f. Determining the derivative of an expression or the derived function by evaluating the limit is called differentiation by ﬁrst principles.

Example 5

For f (x) = x2 + 2x ﬁnd f (x)byﬁrst principles.

Solution f (x + h) − f (x) f (x) = lim h→0 h (x + h)2 + 2(x + h) − (x2 + 2x) = lim h→0 h x2 + 2xh + h2 + 2x + 2h − x2 − 2x = lim h→0 h 2xh + h2 + 2h = lim h→0 h = lim 2x + h + 2 SAMPLEh→0 = 2x + 2 ∴ f (x) = 2x + 2

Chapter 19 — Differentiation of Polynomials 543

Example 6

For f (x) = 2 − x3 ﬁnd f (x)byﬁrst principles.

Solution f (x + h) − f (x) f (x) = lim h→0 h 2 − (x + h)3 − (2 − x3) = lim h→0 h 2 − (x3 + 3x2h + 3xh2 + h3) − (2 − x3) = lim h→0 h −3x2h − 3xh2 − h3 = lim h→0 h = lim −3x2 − 3xh − h2 h→0 =−3x2

Using the TI-Nspire Deﬁne f (x) = 2 − x3. Calculate the gradient of the chord, f (x + h) − f (x) . h Select Limit from the Calculus menu (b 43) and complete as shown.

Using the Casio ClassPad It is a good idea to select Edit—Clear all variables at this point as we will be storing information as variables and a clean variable list SAMPLEis advisable.

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In select Interactive—Deﬁne and enter the information, f (x) = 2 − x3,asshown. Note: Select f from abc keyboard and x from 2D VA R keyboard.

Enter the expression (f(x + h) − f(x))/h as shown and click EXE to produce the algebraic expression. Copy and paste the answer to the next entry line, then highlight it and select Interactive—Transformation—simplify to produce the simpliﬁed expression. Copy and paste this answer to the next entry line, then highlight it and select Interactive—Calculation—lim and set the variable as h, Point as 0 and Direction as 0.

The following results have been obtained: For f (x) = x, f (x) = 1. For f (x) = x2, f (x) = 2x. For f (x) = x3, f (x) = 3x2. For f (x) = x4, f (x) = 4x3. For f (x) = 1, f (x) = 0. This suggests the following general result: For f (x) = xn, f (x) = nxn−1, n = 1, 2, 3,... and for f (x) = 1, f (x) = 0 From the previous section it can be seen that for k,aconstant:

If f (x) = kxn, the derivative function f has rule f (x) = knxn−1. SAMPLEIt is worth making a special note of the results: If g(x) = kf(x), where k is a constant, then g (x) = kf (x).

Chapter 19 — Differentiation of Polynomials 545

That is, the derivative of a number multiple is the multiple of the derivative. Forexample, for g(x) = 5x2, the derived function g (x) = 5(2x) = 10x. Another important rule for differentiation is:

If f (x) = g(x) + h(x), then f (x) = g (x) + h (x).

That is, the derivative of the sum is the sum of the derivatives. Forexample, for f (x) = x2 + 2x, the derived function f (x) = 2x + 2. The process of ﬁnding the derivative function is called differentiation.

Example 7

Find the derivative of x5 − 2x3 + 2, i.e. differentiate x5 − 2x3 + 2 with respect to x.

Solution f (x) = x5 − 2x3 + 2 then f (x) = 5x4 − 2(3x2) + 2(0) = 5x4 − 6x2

Example 8

Find the derivative of f (x) = 3x3 − 6x2 + 1 and f (1).

Solution f (x) = 3x3 − 6x2 + 1 then f (x) = 3(3x2) − 6(2x) + 1(0) = 9x2 − 12x f (1) = 9 − 12 =−3

Using the TI-Nspire Solutions to Example 7 and 8 Select Derivative from the Calculus menu SAMPLE(b 41) and complete as shown.

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Deﬁne f (x) = 3x3 − 6x2 + 1. Select Derivative from the Calculus menu (b 5 1) and complete as shown.

Using the Casio ClassPad Solutions to Examples 7 and 8 Enter and highlight the expression, then from the keyboard select 2D—Calc— and enter x as the variable.

In select Action—Command—Define and enter f (x) = 3x3 − 6x2 + 1asshown (f is from the abc menu in and x is from the grey x keyonthe calculator keyboard). Select Interactive—Calculation—diff and enter the information shown. Then click EXE. In Action—Command—Define define function g(x) = 9x2 − 12x, the derivative function for f(x). Now enter g(1). SAMPLE

Chapter 19 — Differentiation of Polynomials 547

Example 9

Find the gradient of the curve determined by the rule f (x) = 3x3 − 6x2 + 1atthe point (1, −2).

Solution Now f (x) = 9x2 − 12x and f (1) = 9 − 12 =−3. The gradient of the curve is −3atthe point (1, −2).

An alternative notation for the derivative is the following: dy dy If y = x3, then the derivative can be denoted by , so that = 3x2. dx dx

dy In general, if y is a function of x, the derivative of y with respect to x is denoted by and dx with the use of different symbols z,where z is a function of t. The derivative of z with respect dz to t is . y dt In this notation d is not a factor and cannot be cancelled. P This came about because in the eighteenth century the δ standard diagram for ﬁnding the limiting gradient was y labelled as in the ﬁgure. ( is the lower case Greek letter Q δx for ‘d’, and is pronounced delta.) x means a difference in x. x 0 y means a difference in y.

Example 10

dy dx 1 dz a If y = t2, find . b If x = t3 + t, find . c If z = x3 + x2, find . dt dt 3 dx Solution 1 a y = t2 b x = t3 + t c z = x3 + x2 3 dy dx then = 2t then = 3t2 + 1 dz dt dt then = x2 + 2x SAMPLEdx

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Example 11

dy dz a For y = (x + 3)2, find . b For z = (2t − 1)2(t + 2), find . dx dt x2 + 3x dy c For y = , find . d Differentiate y = 2x3 − 1 with respect to x. x dx Solution a It is first necessary to write b Expanding: = + 2 y (x 3) in expanded form. z = (4t2 − 4t + 1)(t + 2) ∴ = 2 + + y x 6x 9 = 4t 3 − 4t2 + t + 8t2 − 8t + 2 dy and = 2x + 6 = 4t3 + 4t2 − 7t + 2 dx dz and = 12t2 + 8t − 7 dt c First divide by x: d y = 2x3 − 1 dy y = x + 3 ∴ = 6x2 dx dy ∴ = 1 dx

Operator notation d ‘Find the derivative of 2x2 − 4x with respect to x’ can also be written as (2x2 − 4x), and, in dx d general, ( f (x)) = f (x). dx

Example 12

Find: d d d a (5x − 4x3) b (5z2 − 4z) c (6z3 − 4z2) dx dz dz

Solution d d d 2 3 − 2 a (5x − 4x3) b (5z − 4z) c (6z 4z ) dx dz dz = − = 2 − SAMPLE= 5 − 12x2 10z 4 18z 8z

Chapter 19 — Differentiation of Polynomials 549

Example 13

Find the coordinates of the points on curves determined by each of the following equations at which the gradient has the given value: a y = x3;gradient = 8 b y = x2 − 4x + 2; gradient = 0 c y = 4 − x3;gradient =−6

Solution dy dy a y = x3 implies = 3x2 b y = x2 − 4x + 2 implies = 2x − 4 dx dx ∴ 3x2 = 8 ∴ 2x − 4 = 0 √ ∴ = 8 ±2 6 x 2 ∴ x =± = 3 3 ∴ coordinates are (2, −2) ∴ coordinates are √ √ √ √ 2 6 16 6 −2 6 −16 6 , and , 3 9 3 9 dy c y = 4 − x3 implies =−3x2 dx ∴ −3x2 =−6 ∴ x2 = 2 √ ∴ x =± 2. 1 3 1 3 ∴ coordinates are 2 2 , 4 − 2 2 and −2 2 , 4 + 2 2

Using the TI-Nspire Deﬁne f (x) = 4 − x3. Take the Derivative and store as df(x). Solve the equation df(x) =−6. Substitute in f (x)toﬁnd the y-coordinates. SAMPLE

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Using the Casio ClassPad In Interactive—Deﬁne, deﬁne the function f (x) = 4 − x3. Select Interactive— Calculate—diff and enter as shown.

Enter and highlight the equation −3x2 =−6 for x.(Yo u can copy the answer from the line above to save re-typing.)√ √ Now enter f ( 2) and f (− 2) to ﬁnd the required y-values.

Exercise 19B

f (x + h) − f (x) Examples 5, 6 1 For each of the following, ﬁnd f (x)byﬁnding lim : h→0 h a f (x) = 3x2 b f (x) = 4x c f (x) = 3 d f (x) = 3x2 + 4x + 3 e f (x) = 2x3 − 4 f f (x) = 4x2 − 5x g f (x) = 3 − 2x + x2

Example 7 2 Find the derivative of each of the following with respect to x. a x2 + 4x b 2x + 1 c x3 − x 1 d x2 − 3x + 4 e 5x3 + 3x2 f −x3 + 2x2 2 3 For each of the following ﬁnd f (x): a f (x) = x12 b f (x) = 3x7 c f (x) = 5x d f (x) = 5x + 3 e f (x) = 3 f f (x) = 5x2 − 3x

5 4 1 1 g f (x) = 10x + 3x h f (x) = 2x4 − x3 − x2 + 2 3 4 dy Example 10 4 For each of the following, ﬁnd : dx SAMPLEa y =−x b y = 10 c y = 4x3 − 3x + 2 1 d y = (x3 − 3x + 6) e y = (x + 1)(x + 2) f y = 2x(3x2 − 4) 3 10x5 + 3x4 g y = , x = 0 2x2

Chapter 19 — Differentiation of Polynomials 551

Example 9 5 a For the curve with equation y = x3 + 1 find the gradient at points: i (1, 2) ii (a, a3 + 1) b Find the derivative of x3 + 1 with respect to x. dy dy 6 a Given that y = x3 − 3x2 + 3x, find . Hence show that ≥ 0 for all x, and dx dx interpret this in terms of the graph of y = x3 − 3x2 + 3x. x2 + 2x dy b Given that y = , for x = 0, find . x dx c Differentiate y = (3x + 1)2 with respect to x.

7 At the points on the following curves corresponding to the given values of x, ﬁnd the y-coordinate and the gradient. a y = x2 − 2x + 1, x = 2 b y = x2 + x + 1, x = 0 c y = x2 − 2x, x =−1 d y = (x + 2)(x − 4), x = 3

2 3 1 e y = 3x − 2x , x =−2 f y = (4x − 5)2, x = 2 8 a For each of the following, ﬁnd f (x) and f (1), if y = f (x) then ﬁnd the {(x, y): f (x) = 1} i.e. the coordinates of the points where the gradient is 1. 1 1 i y = 2x2 − x ii y = 1 + x + x2 2 3 iii y = x3 + x iv y = x4 − 31x b What is the interpretation of {(x, y): f (x) = 1} in terms of the graphs?

Example 12 9 Find: d d d a (3t2 − 4t) b (4 − x2 + x3) c (5 − 2z2 − z4) dt dx dz d d d d (3y2 − y3) e (2x3 − 4x2) f (9.8t2 − 2t) dy dx dt Example 13 10 Find the coordinates of the points on the curves given by the following equations at which the gradient has the given values: a y = x2;gradient = 8 b y = x3;gradient = 12 c y = x(2 − x); gradient = 2 d y = x2 − 3x + 1; gradient = 0 SAMPLEe y = x3 − 6x2 + 4; gradient =−12 f y = x2 − x3;gradient =−1

552 Essential Mathematical Methods1&2CAS 19.3 Graphs of the derived or gradient function Consider the gradient for different intervals y of the graph of y = g(x) shown opposite. At a point (a, g(a)) of the graph y = g(x) R(b, g( b)) y = g(x) the gradient is g(a). Some of the features of the graph are: a x < b the gradient is positive, x b 0 i.e. g(x) > 0 x = b the gradient is zero, i.e. g(b) = 0 S(a, g(a)) b < x < a the gradient is negative, i.e. g(x) < 0 x = a the gradient is zero, i.e. g(a) = 0 x > a the gradient is positive, i.e. g(x) > 0

Example 14

For the graph of f : R → R, ﬁnd: y a {x: f (x) > 0} b {x: f (x) < 0} c {x: f (x) = 0} (5, 6)

Solution a {x: f (x) > 0} = {x: −1 < x < 5} = (−1, 5) x 0 b {x: f (x) < 0} = {x: x < −1} ∪ {x : x > 5} y = f (x) = (−∞, −1) ∪ (5, ∞) c {x: f (x) = 0} = {−1, 5} (–1, –7)

Example 15

Sketch the graph of y = f (x) for each of the following. (It is impossible to determine all features.) c y b y y a y = f (x) (–1.5, 4) y = f (x) y = f (x)

1 x x –3 –1 0 4 x –1 0 0 2 4 SAMPLE(3, –1) (1, –4)

Chapter 19 — Differentiation of Polynomials 553

Solution

a f (x) > 0 for x > 3 y b f (x) = 1 for all x y < < f (x) 0 for x 3 y = f'(x) f (x) = 0 for x = 3 1 y = f ′(x) x 0 3 x 0 y c f (x) > 0 for x > 1 y = f ′(x) f (x) < 0 for −1.5 < x < 1 f (x) > 0 for x < −1.5 − . = = x f ( 1 5) 0 and f (1) 0 −1.5 0 1

An angle associated with the gradient of a curve at a point The gradient of a curve at a point is the gradient of the tangent at that point. A straight line, the tangent, is associated with each point on the curve. If is the angle a straight line makes with the positive direction of the x-axis, then the gradient, m,ofthe straight line is equal to tan , i.e., tan = m. If = 45◦ then tan = 1 and the gradient is 1. If = 20◦ then the gradient of the straight line is tan 20◦. If = 135◦ then tan =−1 and the gradient is −1.

Example 16

Find the coordinates of the points on the curve with equation y = x2 − 7x + 8atwhich the tangent: a makes an angle of 45◦ with the positive direction of the x-axis b is parallel to the line y =−2x + 6.

Solution dy a = 2x − 7 b Note: y =−2x + 6 has gradient −2 dx 2x − 7 = 1 (tan 45◦ = 1) ∴ 2x − 7 =−2 and 2x = 8 and 2x = 5 5 ∴ x = 4 ∴ x = 2 SAMPLEy = 42 − 7 × 4 + 8 =−4 5 13 coordinates are ,− coordinates are (4, −4) 2 4

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Example 17

The planned path for a flying saucer leaving a planet is defined by the equation 1 2 y = x4 + x3 for x > 0. 4 3 The units are kilometres. (The x-axis is horizontal and the y-axis vertical.) a Find the direction of motion when the x-value is: i 2 ii 3 b Find a point on the flying saucer’s path where the path is inclined at 45◦ to the positive x-axis. (i.e. where the gradient of the path is 1). c Are there any other points on the path which satisfy the situation described in part b?

Solution dy a = x3 + 2x2 dx dy dy i When x = 2, = 8 + 8 = 16 ii When x = 3, = 27 + 18 = 45 dx dx tan−116 = 86.42◦ (to the x-axis) tan−145 = 88.73◦ (to the x-axis)

b, c When the flying saucer is flying at 45◦ to the direction of the x-axis, the gradient of the curve of its path is given by tan 45◦. dy Thus to find the point at which this happens we consider equation = tan 45◦. dx ∴ x3 + 2x2 = 1 ∴ x3 + 2x2 − 1 = 0 ∴ + 2 + − = (x 1)(x x 1) 0 √ −1 ± 5 ∴ x =−1orx = 2 √ −1 + 5 The only acceptable solution is x = (x ≈ 0.62) as the other two 2 possibilities give negative values for x and we are only considering positive values for x. Exercise 19C

dy 1 On which of the following curves is positive for all values of x? dx y SAMPLEa y b c y

x x x 0 0 0

Chapter 19 — Differentiation of Polynomials 555

d y e y

x x 0 0

dy 2 On which of the following curves is negative for all values of x? dx a y b y

x x 0 0

c y d y

x 0

e y f y

x 0 x 0

3 For the function f (x) = 2(x − 1)2 ﬁnd the values of x for which: a f (x) = 0 b f (x) = 0 c f (x) > 0 d f (x) < 0 e f (x) =−2 y Example 15 4 For the graph of y = h(x) shown here ﬁnd: –3, 5 1 (4, 6) a {x: h(x) > 0} 2 b {x: h(x) < 0} c {x: h(x) = 0} SAMPLEx 0

1 , –3 2

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5 Which of the graphs labelled A–F correspond to each of the graphs labelled a–f?

a y d y A D dy dy dx dx 0 x x x 0 0 0 x

b y e y B E dy dy dx dx x

x 0 0 0 x 0 x

c y f y C F dy dy dx dx

x x x 0 0 0 x 0

y Example 15 6 For the graph of y = f (x) ﬁnd: a {x: f (x) > 0} (1.5, 3) b {x: f (x) < 0} c {x: f (x) = 0} x 0

(–1, –2)

Example 16 7 Sketch the graph of y = f (x) for each of the following abcdy y y y (3, 4) (3, 4) y = f (x) y = f (x) 1 3 x x x x 0 15 0 1 0 0

SAMPLE(–1, –3) Example 17 8 Find the coordinates of the points on the curve y = x2 − 5x + 6atwhich the tangent: a makes an angle of 45◦ with the positive direction of the x-axis, i.e. where the gradient is 1 b is parallel to the line y = 3x + 4.

Chapter 19 — Differentiation of Polynomials 557

Example 17 9 Find the coordinates of the points on the parabola y = x2 − x − 6atwhich: a the gradient is zero b the tangent is parallel to the line x + y = 6.

10 Use a calculator to plot the graph of y = f (x)where: a f (x) = sin x b f (x) = cos x c f (x) = 2x

11 A car moves away from a set of trafﬁc lights so that the distance S(t) metres covered after t seconds is modelled by S(t) = (0.2)t3. a Find its speed after t seconds. b What will its speed be when t = 1, 3, 5?

12 A rocket is launched from Cape York Peninsula so that after t seconds its height = 2 ≤ ≤ 1 h(t) metres is given by h(t) 20t ,0 t 150. After 2 2 minutes this model is no longer appropriate. a Find the height and the speed of the rocket when t = 150. b After how long will its speed be 1000 m/s?

13 The curve with equation y = ax2 + bx has a gradient of 3 at the point (2, −2). a Find the values of a and b. b Find the coordinates of the point where the gradient is 0. 19.4 Limits and continuity Limits We consider the limit of a function f (x)tobethe value that f (x) approaches as x approaches a given value. lim f (x) = p means that, as x approaches a, f (x) approaches p.Itisimportant to x→a understand that it is possible to get as close as desired to p as x approaches a. Note that f (x) may or may not be deﬁned at x = a. For many functions f (a)isdeﬁned, so to evaluate the limit we simply substitute the value a into the rule for the function.

Example 18

If f (x) = 3x2 ﬁnd lim 3x2. x→2 Solution Since f (x) = 3x2 is deﬁned at x = 2

lim 3x2 = 3(2)2 x→2 SAMPLE= 12 If the function is not deﬁned at the value for which the limit is to be found, a different procedure is used.

558 Essential Mathematical Methods1&2CAS

Example 19

2x2 − 5x + 2 For f (x) = , x = 2, ﬁnd lim f (x). x − 2 x→2 Solution Observe that f (x)isdeﬁned for x ∈ R \ {2}. Examine the behaviour of f (x) for values of x ‘near’ 2.

x < 2 x > 2

f (1.7) = 2.4 f (2.3) = 3.6 y f (1.8) = 2.6 f (2.2) = 3.4 f (1.9) = 2.8 f (2.1) = 3.2 3 f f (1.99) = 2.98 f (2.01) = 3.02 f (1.999) = 2.998 f (2.001) = 3.002

From the table it is apparent that as x takes values closer and closer to 2, regardless of whether x approaches 2 from the left or from the right, the x 0 1 2 values of f (x) become closer and closer to 3. i.e. lim f (x) = 3 –1 x→2 This may also be seen by observing that: (2x − 1)(x − 2) f (x) = , n = 2 x − 2 = 2x − 1, x = 2,

The graph of f : R \ {2} → R, f (x) = 2x − 1isshown.

The following important results are useful for the evaluation of limits: lim ( f (x) + g(x)) = lim f (x) + lim g(x) x→c x→c x→c i.e. the limit of the sum is the sum of the limits. lim kf(x) = k lim f (x), k being a given number x→c x→c lim ( f (x)g(x)) = lim f (x) lim g(x) x→c x→c x→c i.e. the limit of the product is the product of the limits. lim f (x) f (x) → lim = x c , provided lim g(x) = 0 x→c g(x) lim g(x) x→c SAMPLEx→c i.e. the limit of the quotient is the quotient of the limits.

Chapter 19 — Differentiation of Polynomials 559

Example 20

Find: x2 − 3x x2 − x − 2 a lim(x2 + 2) b lim c lim x→0 x→3 x − 3 x→2 x − 2 x2 − 7x + 10 d lim(2x + 1)(3x − 2) e lim x→3 x→3 x2 − 25 Solution a lim(x2 + 2) = lim x2 + lim 2 = 0 + 2 = 2 x→0 x→0 x→0 x2 − 3x x(x − 3) b lim = lim = lim x = 3 x→3 x − 3 x→3 x − 3 x→3 x2 − x − 2 (x − 2)(x + 1) c lim = lim = lim(x + 1) = 3 x→2 x − 2 x→2 x − 2 x→2 d lim(2x + 1)(3x − 2) = lim(2x + 1) lim(3x − 2) = 7 × 7 = 49 x→3 x→3 x→3 2 lim(x − 2) x − 7x + 10 (x − 2)(x − 5) → 1 e lim = lim = x 3 = x→3 x2 − 25 x→3 (x + 5)(x − 5) lim(x + 5) 8 x→3

Example 21

Find: 5x + 2 a lim(3h + 4) b lim 4x(x + 2) c lim h→0 x→2 x→3 x − 2 Solution a lim(3h + 4) = lim(3h) + lim(4) h→0 h→0 h→0 = 0 + 4 = 4

b lim 4x(x + 2) = lim(4x) lim(x + 2) x→2 x→2 x→2 = 8 × 4 = 32 5x + 2 c lim = lim(5x + 2) ÷ lim(x − 2) x→3 x − 2 x→3 x→3 = 17 ÷ 1 = 17

The notation of limits is used to describe the behaviour of graphs, and a similar notation has been used previously in the book. SAMPLE1 Consider f : R \{0}→R, f (x) = . Observe that as x → 0, both from the left and from x2 the right, f (x) increases without bound. The limit notation for this is lim f (x) =∞. x→0

560 Essential Mathematical Methods1&2CAS

1 For g: R \{0}→R, g(x) = , the behaviour of g(x)asx approaches 0 from the left is x different from the behaviour as x approaches 0 from the right. With limit notation this is written as:

lim g(x) =−∞ and lim f (x) =∞ x→0− x→0+ Now examine this function as the magnitude of x becomes y very large. It can be seen that as x increases without bound through positive values, the corresponding values of g(x) approach zero. Likewise as x decreases 1 g(x) = x without bound through negative values, the x corresponding values of g(x) also approach zero. 0 Symbolically this is written as:

lim g(x) = 0 and lim g(x) = 0 x→∞ x→−∞ Many functions approach a limiting value or limit as x approaches ±∞. Left and right limits An idea which is useful in the following discussion is the existence of limits from the left and from the right. If the value of f (x) approaches the number p as x approaches a from the right-hand side, then it is written as lim f (x) = p. If the value of f (x) approaches the number p as x + x→a approaches a from the left-hand side, then it is written as lim f (x) = p. − x→a The limit as x approaches a exists only if the limits from the left and the right both exist and are equal. Then lim f (x) = p. x→a The following is an example of when the limit does not exist for a particular value.   x3 for 0 ≤ x < 1 = = Let f (x)  5 for x 1 6 for 1 < x ≤ 2

It is clear from the graph of f that the y lim f (x) does not exist. However, if x is x→1 6 allowed to approach 1 from the left, then 5 f (x) approaches 1. On the other hand if x is allowed to approach 1 from the right, then 4 f (x) approaches 6. Also note that f (1) = 5. 3 2 SAMPLE1 x 0 1 2

Chapter 19 — Differentiation of Polynomials 561 Continuity at a point: informal definition A function with rule f (x)issaid to be continuous when x = a if the graph of y = f (x) can be drawn through the point with coordinates (a, f (a)) without a break. Otherwise there is said to be a discontinuity at x = a. Most of the functions considered in this course are continuous for their domains. A more formal deﬁnition of continuity follows. A function f is continuous at a point a if f (a), lim f (x) and lim f (x) all exist and + − x→a x→a are equal. Or equivalently:

A function f is continuous at the point x = a if the following three conditions are met: 1 f (x)isdeﬁned at x = a 2 lim f (x)exists 3 lim f (x) = f (a) x→a x→a

The function is said to be discontinuous at a point if it is not continuous at that point. A function is said to be continuous everywhere if it is continuous for all real numbers. 1 The polynomial functions are all continuous for R. The function with rule f (x) = has a x discontinuity at x = 0, as f (0) is not deﬁned. It is continuous elsewhere in its domain. Hybrid functions, as introduced in Chapter 6, provide examples of functions which have points of discontinuity where the function is deﬁned.

Example 22

State the values for x for which the functions whose graphs are shown below have a discontinuity. c y a y b y 3 3 3 2 2 2 1 x 1 1 –1 0 1 x x –1 0 1 –1 0 1

Solution a There is a discontinuity at x = 1, as f (1) = 3but lim f (x) = lim f (x) = 2. + − x→1 x→1 b There is a discontinuity at x =−1, as f (−1) = 2 and lim f (x) = 2but − x→−1 lim f (x)=−∞and a discontinuity at x = 1as f (1) = 2 and lim f (x) = 2 SAMPLE+ − x→−1 x→1 but lim f (x) = 3. + x→1 c There is a discontinuity at x = 1, as f (1) = 1 and lim f (x) = 1but − x→1 lim f (x) = 2. + x→1

562 Essential Mathematical Methods1&2CAS

Example 23

For each of the following functions state the values of x for which there is a discontinuity and use the deﬁnition of continuity in terms of f (a), lim f (x) and lim f (x)toexplain why each + − x→a x→a is discontinuous: 2x if x ≥ 0 x2 if x ≥ 0 a f (x) = b f (x) = − + < − + <  2x 1ifx 0 2x 1ifx 0

x if x ≤−1 x2 + 1ifx ≥ 0 c f (x) = x2 if −1 < x < 0 d f (x) =  −2x + 1ifx < 0 −2x + 1ifx ≥ 0

x if x ≥ 0 e f (x) = −2x if x < 0

Solution a f (0) = 0but lim f (x) = 1, therefore there is a discontinuity at x = 0 − x→0 b f (0) = 0but lim f (x) = 1, therefore there is a discontinuity at x = 0 − x→0 c f (−1) =−1but lim f (x) = 1, therefore there is a discontinuity at x = 1 x→−1+ f (0) = 1butlim f (x) = 0, therefore there is a discontinuity at x = 0 x→0− d No discontinuity e No discontinuity

Exercise 19D

Examples 18–21 1 Find the following limits: a lim 15 − − → b lim(x 5) c lim (3x 5) x 3 x→6 1 x→ 2 (t − 2) t2 + 2t + 1 (x + 2)2 − 4 d lim e lim f lim t→−3 (t + 5) t→−1 t + 1 x→0 x t2 − 1 √ 2 − g lim + x 2x → − h lim x 3 i lim t 1 t 1 x→9 x→0 x x3 − 8 2 − − 2 − + j lim 3x x 10 x 3x 2 → − k lim l lim x 2 x 2 x→2 x2 + 5x − 14 x→1 x2 − 6x + 5 Example 22 2 For each of the following graphs give the values of x at which a discontinuity occurs. Give reasons.

SAMPLEa y b y

6 2 x x 134 2 5 7

Chapter 19 — Differentiation of Polynomials 563

Example 23 3 For each of the following functions state the values of x at which there is a discontinuity and use the deﬁnition of continuity in terms of f (a), lim f (x) and lim f (x)toexplain + − x→a x→a why each stated value of x corresponds to a discontinuity. 3x if x ≥ 0 x2 + 2ifx ≥ 1 a f (x) = b f (x) = −2x + 2ifx < 0 −2x + 1ifx < 1  −x if x ≤−1 c f (x) = x2 if −1 < x < 0  −3x + 1ifx ≥ 0

4 The rule of a particular function is given below. For what values of x is the graph of this function discontinuous?  2, x < 1 y = (x − 4)2 − 9, 1 ≤ x < 7  x − 7, x ≥ 7

19.5 When is a function differentiable? f (x + h) − f (x) A function f is said to be differentiable at x if lim exists. h→0 h The polynomial functions considered in this chapter are differentiable for all x.However this is not true for all functions.

x if x ≥ 0 Let f : R → R, f (x) = −x if x < 0

This function is called the modulus function or absolute value function and it is denoted by f (x) =|x|, f is not differentiable at x = 0:   h y + −  h > 0 f (0 h) f (0) = h  −h y = x h  h < 0 h

> = 1 h 0 −1 h < 0

f (0 + h) − f (0) x So lim does not exist. 0 h→0 h i.e. f is not differentiable at x = 0. SAMPLENote: The gradient to the left of 0 is −1 and to the right of 0 the gradient is 1.

564 Essential Mathematical Methods1&2CAS

Example 24

y x if x ≥ 0 Let f : R → R, f (x) = (=|x|) −x if x < 0 Sketch the graph of the derivative for a suitable domain. 1

Solution x 1ifx > 0 0 f (x) = −1ifx < 0 –1 f (x)isnot deﬁned at x = 0

Example 25 y 2 Draw a sketch graph of f where the graph of f is as illustrated. Indicate where f is not deﬁned. x 0–1 1 Solution y The derivative does not exist at x = 0; i.e. the function is not differentiable at x = 0. 2

x –1 0 1 –2

It was shown in the previous section that some hybrid functions are continuous for R. There are hybrid functions which are differentiable for R. The smoothness of the ‘joins’ determines if this is the case.

Example 26

For the function with following rule ﬁnd f (x) and sketch the graph of y = f (x):

x2 + 2x + 1ifx ≥ 0 f (x) = 2x + 1ifx < 0 y

Solution 2x + 2ifx ≥ 0 f (x) = SAMPLE2ifx < 0 2 In particular f (0) is deﬁned and is equal x 0 to 2. Also f (0) = 1. The two sections of the graph of y = f (x) join smoothly at (0, 1).

Chapter 19 — Differentiation of Polynomials 565

Example 27

For the function with rule

x2 + 2x + 1ifx ≥ 0 f (x) = x + 1ifx < 0

state the set of values for which the derivative is deﬁned, ﬁnd f (x) for this set of values and sketch the graph of y = f (x). y Solution 2x + 2ifx > 0 f (x) = 1ifx < 0

2 f (0) is not deﬁned as the limits from 1 the left and right are not equal. The x function is differentiable for R\{0}. 0

Exercise 19E

Examples 24–26 1 In each of the ﬁgures below a function graph f is given. Sketch the graph of f .Obviously your sketch of f cannot be exact, but f (x) should be 0 at values of x for which the gradient of f is zero; f (x) should be < 0where the original graph slopes downward, and so on. a y b y c y

f f x x x –1 0 1 –4 0 42–2 –1 0 1 f

d y e y f y

f f f x x x 0 –1 0 1 –1 0 1 –1 1

Example 26 2 For the function with following rule ﬁnd f (x) and sketch the graph of y = f (x):

−x2 + 3x + 1ifx ≥ 0 f (x) = SAMPLE3x + 1ifx < 0

566 Essential Mathematical Methods1&2CAS

Example 27 3 For the function with following rule state the set of values for which the derivative is deﬁned, ﬁnd f (x) for this set of values and sketch the graph of y = f (x):

x2 + 2x + 1ifx ≥ 1 f (x) = −2x + 3ifx < 1

Example 27 4 For the function with following rule state the set of values for which the derivative is deﬁned, ﬁnd f (x) for this set of values and sketch the graph of y = f (x):

−x2 − 3x + 1ifx ≥−1 f (x) = −2x + 3ifx < −1

SAMPLE

Chapter 19 — Differentiation of Polynomials 567 Review Chapter summary

The notation for limit as h approaches 0 is written as lim. h→0 The following are important results for limits: r lim ( f (x) + g(x)) = lim f (x) + lim g(x) x→c x→c x→c i.e. the limit of the sum is the sum of the limits. r lim kf(x) = k lim f (x), k being a given number x→c x→c r lim ( f (x)g(x)) = lim f (x) lim g(x) x→c x→c x→c i.e. the limit of the product is the product of the limits. lim f (x) f (x) → r lim = x c , provided lim g(x) = 0 x→c g(x) lim g(x) x→c x→c i.e. the limit of the quotient is the quotient of the limits. A function f is deﬁned as continuous at the point x = a if three conditions are met: 1 f (x)isdeﬁned at x = a 2 lim f (x)exists x→a 3 lim f (x) = f (a) x→a The function is said to be discontinuous at a point if it is not continuous at that point. We say that a function is continuous everywhere if it is continuous for all real numbers. For the graph of y = f (x)ofthe function f : R → R: y f (x + h) − f (x) The gradient of the chord PQ = . h Q (x + h, f (x+h)) The gradient of the graph at P is given by f (x + h) − f (x) lim . P (x, f (x)) h→0 h x 0 This limit gives a rule for the derived function denoted by f ,where f : R → R and f (x + h) − f (x) f (x) = lim h→0 h The general rule of the derived function of f (x) = xn, n = 1, 2, 3,... For f (x) = xn, f x = nxn−1, n = 1, 2, 3,... For f (x) = 1, f (x) = 0 Forexample: For f (x) = x2, f (x) = 2x.Forf (x) = x3, f (x) = 3x2. For f (x) = x4, f (x) = 4x3.Forf (x) = 1, f (x) = 0. SAMPLEThe derivative of a number multiple is the multiple of the derivative: For g(x) = kf(x), where k is a constant g(x) = kf (x) Forexample: g(x) = 3x2, g(x) = 3(2x) = 6x

568 Essential Mathematical Methods1&2CAS

The derivative of a constant is always zero: For g(x) = a, g(x) = 0 Forexample: f (x) = 27.3, f (x) = 0 The derivative of the sum is the sum of the derivatives: If f (x) = g(x) + h(x) then = + Review f (x) g (x) h (x). Forexample: f (x) = x2 + x3, f (x) = 2x + 3x2 g(x) = 3x2 + 43, g(x) = 3(2x) + 4(3x2) = 6x + 12x2

At a point (a, g(a)) on the curve y = g(x) the gradient is g(a). y For x < b the gradient is positive, i.e. g(x) > 0 R(b, g(b)) x = b the gradient is zero, i.e. g(b) = 0 y = g(x) b < x < a the gradient is negative, i.e. g(a) < 0 a x x = a the gradient is zero, i.e. g (a) = 0 b 0 x > a the gradient is positive, i.e. g(x) > 0 S(a, g(a))

Multiple-choice questions

1 The gradient of the curve y = x3 + 4x at the point where x = 2is A 12 B 4 C 10 D 16 E 8 2 The gradient of the chord of the curve y = 2x2 between the points where x = 1 and x = 1 + h is given by A 2(x + h)2 − 2x2 B 4 + 2h C 4 D 4x E 4 + h dy 3 If y = 2x4 − 5x3 + 2, then equals dx A 8x3 − 5x2 + 2 B 4x4 − 15x2 + 2 C 4x4 − 10x2 D 8x3 − 15x + 2 E 8x3 − 15x2 4 If f (x) = x2(x + 1) then f (−1) equals A −1 B 1 C 2 D −2 E 5 5 If f (x) = (x − 3)2, then f (x) equals A x − 3 B x − 6 C 2x − 6 D 2x + 9 E 2x 2x4 + 9x2 dy 6 If y = , then equals 3x dx 2x4 8x3 + 18x A + 6x B 2x + 3 C 2x2 + 3 D E 8x3 + 18x 3 dy 3 SAMPLE= 2 − + ≥ 7 Given that y x 6x 9, the values of x for which 0 are dx A x ≥ 3 B x > 3 C x ≥−3 D x ≤−3 E x < 3 8 If y = 2x4 − 36x2, the points at which the tangent to the curve is parallel to the x-axis are A 1, 0 and 3 B 0 and 3 C −3 and 3 D 0 and −3 E −3, 0 and 3

Chapter 19 — Differentiation of Polynomials 569 Review

9 The coordinates of the point on the graph of y = x2 + 6x − 5atwhich the tangent is parallel to the line y = 4x are A (−1, − 10) B (−1, −2) C (1, 2) D (−1, 4) E (−1, 10) dy 10 If y =−2x3 + 3x2 − x + 1, then is equals dx A 6x2 + 6x − 1 B −6x2 + 6x C −6x2 + 3x − 1 D −6x2 + 6x − 1 E 6x2 − 6x − 1

Short-answer questions (technology-free)

dy 1 Find when: dx a y = 3x2 − 2x + 6 b y = 5 c y = 2x(2 − x) d y = 4(2x − 1)(5x + 2) e y = (x + 1)(3x − 2) f y = (x + 1)(2 − 3x) dy 2 Find when: dx (x + 3)(2x + 1) a y =−x b y = 10 c y = 4 2x3 − x2 x4 + 3x2 d y = e y = 3x 2x2 3 For each of the following functions ﬁnd the y-coordinates and the gradient at the point on the curve for the given value of x: a y = x2 − 2x + 1, x = 2 b y = x2 − 2x, x =−1 c y = (x + 2)(x − 4), x = 3 d y = 3x2 − 2x3, x =−2 4 Find the coordinates of the points on the curves given by the following equations at which the gradient has the given value: dy dy a y = x2 − 3x + 1; = 0 b y = x3 − 6x2 + 4; =−12 dx dx dy dy c y = x2 − x3; =−1 d y = x3 − 2x + 7; = 1 dx dx dy dy e y = x4 − 2x3 + 1; = 0 f y = x(x − 3)2 ; = 0 dx dx 5 For the function with rule f (x) = 3(2x −1)2 ﬁnd the values of x for which: a f (x) = 0 b f (x) = 0 c f (x) > 0 d f (x) < 0 e f (x) > 0 f f (x) = 3 6 The curve with equation y = ax2 + bx has a gradient of 3 at the point (1, 1). Find: a the values of a and b b the coordinates of the points where the gradient is 0. 7 Sketch the graph of y = f (x). (All details cannot be y SAMPLEdetermined but the axis intercepts and shape of graph 5 can be determined.) –1 2 x –1

570 Essential Mathematical Methods1&2CAS

= y 8 For the graph of y h(x) ﬁnd: (4, 3) a {x: h (x) > 0} y = h(x) x b {x: h (x) < 0} 0 c {x: h (x) = 0} (–1, –4)

Review Extended-response questions

1 The diagram to the right shows part of the graph dy of against x. dy dx Sketch a possible shape of y against x over dx the same interval if: r y =−1when x =−1 –1 0 25x r y = 0when x = 0 r y = 1when x = 2. 2 The graph shown is that of a polynomial of the form y P(x) = ax3 + bx2 + cx + d. R(–2, 3) Find the values of a, b, c and d. Note: Q(1, −2) is not a turning point. x

Q(1, –2) 1 1 3 A body moves in a path described by the equation y = x5 + x4, x ≥ 0. 5 2 Units are in kilometres and x and y are the horizontal and vertical axes respectively. a What will be the direction of motion (give the answer as angle between direction of motion and the x-axis) when the x-value is i 1 km? ii 3 km? b Find a value of x for which the gradient of the path is 32. 4 A trail over a mountain pass can be modelled by the curve with equation y = 2 + 0.12x − 0.01x3,where x and y are, respectively, the horizontal and vertical distances measured in kilometres, 0 ≤ x ≤ 3. a Find the gradients at the beginning and the end of the trail. b Calculate the point where the gradient is zero, and calculate also the height of the pass. 5 A tadpole begins to swim vertically upwards in a pond and after t seconds it is 25 − 0.1t3 cm below the surface. a How long does the tadpole take to reach the surface, and what is its speed then? SAMPLEb What is the average speed over this time? 6aShow that the gradients of the curve y = x(x − 2) at the points (0, 0) and (2, 0) only differ in sign. What is the geometrical interpretation for this? b If the gradient of the curve y = x(x − 2)(x − 5) at the points (0, 0), (2, 0) and (5, 0) are 1 1 1 l, m and n respectively, show that + + = 0. l m n

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