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Mathematics

UNIT 1 OUTCOME 3 Differentiation

Contents Differentiation 33 1 Introduction to Differentiation 33 2 Finding the Derivative 34 3 Differentiating with Respect to Other Variables 38 4 Rates of Change 39 5 Equations of Tangents 40 6 Increasing and Decreasing Curves 44 7 Stationary Points 45 8 Determining the Nature of Stationary Points 46 9 Curve Sketching 49 10 Closed Intervals 51 11 Graphs of Derivatives 53 12 Optimisation 54

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OUTCOME 3 Differentiation

1 Introduction to Differentiation From our work on Straight Lines, we saw that the gradient (or “steepness”) of a line is constant. However, the “steepness” of other curves may not be the same at all points. In order to measure the “steepness” of other curves, we can use lines which give an increasingly good approximation to the curve at a particular point. On the curve with equation y= f( x ) , y y= f( x ) suppose point A has coordinates (a, f( a )) . f( a+ h ) B At the point B where x= a + h , we have y= fa( + h ). A f( a ) Thus the chord AB has gradient x O a a+ h fa( + h) − fa( ) m = y AB + − y= f( x ) a h a fa()+ h − fa() = . f( a+ h ) B h A If we let h get smaller and smaller, i.e. h → 0 , f( a ) then B moves closer to A. This means that O a a+ h x mAB gives a better estimate of the “steepness” of the curve at the point A.

We use the notation f′( a ) for the “steepness” of the curve when x= a . So fa( + h) − fa( ) f′() a = lim . h→0 h Given a curve with equation y= f( x ) , an expression for f′( x ) is called the derivative and the process of finding this is called differentiation . It is possible to use this definition directly to find derivates, but you will not be expected to do this. Instead, we will learn rules which allow us to quickly find derivatives for certain curves.

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2 Finding the Derivative

The basic rule for differentiating f( x) = x n , n ∈ ℝ , with respect to x is: If fx()()= xn then fxnx′ = n −1 . Stated simply: the power ( n) multiplies to the front of the x term, and the power lowers by one (giving n −1).

EXAMPLES 1. Given f() x= x 4 , find f′( x ). f′( x) = 4 x 3 .

2. Differentiate f() x= x −3 , x ≠ 0, with respect to x. f′( x) = − 3 x −4 .

For an expression of the form y =… , we denote the derivative with respect dy to x by . dx EXAMPLE − 1 3. Differentiate y= x 3 , x ≠ 0 , with respect to x.

dy − 4 = − 1 x 3 . dx 3

When finding the derivative of an expression with respect to x, we use the d notation dx . EXAMPLE 3 4. Find the derivative of x 2 , x ≥ 0 , with respect to x.

3 1 d 2 = 3 2 dx (x) 2 x .

Preparing to differentiate It is important that before you differentiate, all brackets are multiplied out and there are no fractions with an x term in the denominator (bottom line). For example:

− 1 5 − 2 1 −3 3 −2 1 2 1 1 −5 5 3 = x = 3x = x = x 2 = x . x3 x 2 x 4x5 4 4 3 x 4

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EXAMPLES 1. Differentiate x with respect to x, where x > 0.

1 x= x 2

1− 1 d x2= 1 x 2 dx () 2 Note 1 It is good practice to = . tidy up your answer. 2 x 1 dy 2. Given y = , where x ≠ 0, find . x 2 dx y= x −2 dy − = − 2x 3 dx 2 = − . x 3 Terms with a coefficient For any constant a, if fx( ) = a × gx( ) then f′( x) = agx × ′ ( ) . Stated simply: constant coefficients are carried through when differentiating.

So if f( x) = ax n then f′() x= anx n−1 .

EXAMPLES 1. A function f is defined by f( x) = 2 x 3 . Find f′( x ). f′( x) = 6 x 2 .

2. Differentiate y= 4 x −2 with respect to x, where x ≠ 0.

dy − = − 8x 3 dx 8 = − . x 3 2 3. Differentiate , x ≠ 0, with respect to x. x 3 d −3= − − 4 dx (2x) 6 x 6 = − . x 4

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3 dy 4. Given y = , x > 0, find . 2 x dx

− 1 = 3 2 y2 x dy − 3 = − 3 x 2 dx 4 3 = − 3 . 4 x

Differentiating more than one term The following rule allows us to differentiate expressions with several terms. If fx( ) = gx( ) + hx( ) then fx′( ) = gx ′( ) + hx ′ ( ) .

Stated simply: differentiate each term separately.

EXAMPLES 1. A function f is defined for x ∈ℝ by fx( ) =3 x3 − 2 x 2 + 5 x . Find f′( x ).

fx′( ) =9 x2 − 4 x + 5.

2. Differentiate yx=24 − 4 x 3 + 3 x 2 ++ 62 x with respect to x. dy =8x3 − 12 x 2 ++ 6 x 6. dx

Note 1 − 3 The derivative of an x term (e.g. 3x , 2 x , 10 x ) is always a constant. For example: d () = d −1 = − 1 dx 6x 6, dx ( 2x) 2 . The derivative of a constant (e.g. 3, 20, π) is always zero. For example: d ()3= 0, d −1 = dx dx ( 3 ) 0.

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Differentiating more complex expressions We will now consider more complex examples where we will have to use several of the rules we have met.

EXAMPLES 1 1. Differentiate y = , x > 0, with respect to x. 3x x 1 − 3 = = 1 2 y3 3 x 3x 2 dy − 5 =1 × − 3 x 2 dx 3 2 Note You need to be − 5 = − 1 x 2 confident working with 2 indices and fractions. 1 = − 5 . 2 x dy 2. Find when y=( x −3)( x + 2 ) . dx y=( x −3)( x + 2 ) Remember =x2 +2 x − 3 x − 6 Before differentiating, the brackets must be =x2 − x − 6 multiplied out. dy =2x − 1. dx x 1 3. A function f is defined for x ≠ 0 by f() x = + . Find f′( x ). 5 x 2 () =1 + −2 fx5 xx ′() =1 − −3 f x5 2 x 2 =1 − . 5 x 3 x4− 3 x 2 4. Differentiate with respect to x, where x ≠ 0. 5x x4− 3 x 2 x 4 3 x 2 = − 5x 5 x 5 x =1 3 − 3 5x 5 x d 1 3−3 = 3 2 − 3 . dx ()5x 5 x 5 x 5

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x3+3 x 2 − 6 x 5. Differentiate , x > 0, with respect to x. x x32+36 x − xx 32 36 x x =1 + 1 − 1 x x2 x 2 x 2 3−1 2 − 1 1 − 1 =x2 +3 x 2 − 6 x 2 Remember 5 3 1 xa =x2 +3 x 2 − 6 x 2 = xa− b . xb 5 31 3 1− 1 d 222+− =5 2 −9 2 − 2 dx (xxx3 6) 2 x 2 xx 3 3 3 =5 x −9 x − . 2 2 x 6. Find the derivative of y= xx( 2 + 3 x ), x > 0, with respect to x.

1 1 5 5 yxx=2( 2 + x3 ) =+ x 2 x 6 Remember xab x= x ab+ . dy 3 − 1 =5x2 + 5 x 6 dx 2 6 3 5 =5 x + . 2 6 6 x 3 Differentiating with Respect to Other Variables So far we have differentiated functions and expressions with respect to x. However, the rules we have been using still apply if we differentiate with respect to any other variable. When modelling real-life problems we often use appropriate variable names, such as t for time and V for volume.

EXAMPLES 1. Differentiate 3t2 − 2 t with respect to t. d 2 − = − dt (3t 2 t) 6 t 2. 2. Given A( r) = π r 2 , find A′( r ) . A( r) = π r 2 Remember π is just a constant. A′() r= 2π r . When differentiating with respect to a certain variable, all other letters are treated as constants. EXAMPLE Note 3. Differentiate px 2 with respect to p. Since we are differentiating with 2 d ( px2) = x 2 . respect to p, we treat x dp as a constant.

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4 Rates of Change The derivative of a function describes its “rate of change”. This can be evaluated for specific values by substituting them into the derivative. EXAMPLES 1. Given f( x) = 2 x 5 , find the rate of change of f when x = 3. f′( x) =10 x 4 f ′()()3= 10 34 =×= 10 81 810.

1 2. Given y = 2 for x ≠ 0, calculate the rate of change of y when x = 8 . x 3 − 2 = 3 dy 2 y x At x = 8, = − 5 dx 33 8 dy − 5 = − 2 x 3 2 dx 3 = − 3× 2 5 2 = − 5 = − 2 3x 3 96 = − 1 2 48 . = − 5 . 3 3 x Displacement, velocity and acceleration The velocity v of an object is defined as the rate of change of displacement s with respect to time t. That is: ds v = . dt Also, acceleration a is defined as the rate of change of velocity with respect to time: dv a = . dt EXAMPLE 3. A ball is thrown so that its displacement s after t seconds is given by st( ) =23 t − 5 t 2 . Find its velocity after 2 seconds. vt( ) = st′( ) =−23 10t by differentiating sttt() =− 23 52 with respec t to t . Substitute t = 2 into v( t ) : v(2) = 23 − 10( 2) = 3. After 2 seconds, the ball has velocity 3 metres per second.

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5 Equations of Tangents As we already know, the gradient of a straight line is constant. We can determine the gradient of a curve, at a particular point, by considering a straight line which touches the curve at the point. This line is called a tangent .

tangent

The gradient of the tangent to a curve y= f( x ) at x= a is given by f′( a ) . This is the same as finding the rate of change of f at a. To work out the equation of a tangent we use y− b = mx( − a ). Therefore we need to know two things about the tangent: • a point, of which at least one coordinate will be given; • the gradient, which is calculated by differentiating and substituting in the value of x at the required point.

EXAMPLES 1. Find the equation of the tangent to the curve with equation y= x 2 − 3

at the point (2,1 ) .

We know the tangent passes through (2,1 ) . To find its equation, we need the gradient at the point where x = 2 : y= x 2 − 3 dy = 2x dx At x= 2, m =×= 2 2 4.

Now we have the point (2,1 ) and the gradient m = 4 , so we can find the equation of the tangent: y− b = mx( − a ) y−1 = 4() x − 2 y−1 = 4 x − 8 4x− y − 7 = 0.

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2. Find the equation of the tangent to the curve with equation y= x3 − 2 x at the point where x = − 1. We need a point on the tangent. Using the given x-coordinate, we can find the y-coordinate of the point on the curve: y= x3 − 2 x =−()()13 − 2 − 1 = −1 + 2

=1 So the point is () − 1, − 1 . We also need the gradient at the point where x = − 1: y= x3 − 2 x dy =3x 2 − 2 dx At x=− 1, m = 3( − 1)2 −= 2 1.

Now we have the point (−1,1 ) and the gradient m =1, so the equation of the tangent is: y− b = mx( − a ) y−1 = 1() x + 1 x− y +2 = 0. 1 3. A function f is defined for x > 0 by f() x = . x Find the equation of the tangent to the curve y= f( x ) at P. y

P 2 y= f( x ) O x We need a point on the tangent. Using the given y-coordinate, we can find the x-coordinate of the point P: f( x ) = 2 1 = 2 x 1 1 = x 2 So the point is ()2 , 2 .

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= 1 We also need the gradient at the point where x 2 : f( x) = x −1 f′() x= − x −2 1 = − . x 2 1 =1 =− =− At x2 , m 1 4. 4 1 Now we have the point ( 2 , 2 ) and the gradient m = − 4, so the equation of the tangent is: y− b = mx( − a ) − =− − 1 y2 4 () x 2 y−2 =− 4 x + 2 4x+ y − 4 = 0.

2 4. Find the equation of the tangent to the curve y= 3 x at the point where x = − 8. We need a point on the tangent. Using the given x-coordinate, we can work out the y-coordinate: 2 y =3 − 8 =() − 2 2

=4 So the point is () − 8, 4 . We also need the gradient at the point where x = − 8: 2 2 2 y=3 x = x 3 At x= − 8, m = 33 8 dy − 1 = 2 3 2 3 x = dx × 2 3 2 = = 1 3 3 x 3 . 1 − = Now we have the point ( 8, 4 ) and the gradient m 3 , so the equation of the tangent is: y− b = mx( − a ) − =1 () + y43 x 8 3y− 12 = x + 8 x−3 y + 20 = 0.

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=13 − 1 2 ++ 5. A curve has equation y3 x 2 x2 x 5 . Find the coordinates of the points on the curve where the tangent has gradient 4. The derivative gives the gradient of the tangent: dy =x2 − x + 2. dx We want to find where this is equal to 4: x2 − x +2 = 4 Remember Before solving a x2 − x −2 = 0 quadratic equation you ()()x+1 x − 2 = 0 need to rearrange to get “quadratic= 0 ”. x= −1 or x = 2. Now we can find the y-coordinates by using the equation of the curve: =1() −3 − 1 ()() − 2 +−+ =1()3 − 1 ()() 2 + + y 31 2 1 215 y 32 2 2 225 =−1 − 1 − + =8 −4 + + 3 2 2 5 3 2 4 5

= − 5 = + 8 3 6 7 3 = 13 = 29 6 3 . − 13 29 So the points are ( 1, 6 ) and (2, 3 ) .

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6 Increasing and Decreasing Curves dy A curve is said to be strictly increasing when > 0. dx dy This is because when > 0, tangents will slope upwards dx from left to right since their gradients are positive. This means the curve is also “moving upwards”, i.e. strictly increasing. Similarly: dy A curve is said to be strictly decreasing when < 0. dx y

strictly decreasing strictly dy strictly increasing < 0 dx increasing dy dy > 0 > 0 dx dx O x

EXAMPLES 2 1. A curve has equation y=4 x 2 + . x Determine whether the curve is increasing or decreasing at x =10 .

− 1 y=4 x2 + 2 x 2

dy − 3 =8x − x 2 dx 1 =8x − 3 . x dy 1 When x =10 , =8 × 10 − 3 dx 10 Note 1 = − 1 80 <1. 10 10 10 10 > 0. dy Since > 0, the curve is increasing when x =10 . dx

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=1 3 + 2 +− 2. Show that the curve y3 x x x 4 is never decreasing. dy =x2 +2 x + 1 Remember dx The result of squaring =()x + 1 2 any number is always greater than, or equal to, ≥ 0. zero. dy Since is never less than zero, the curve is never decreasing. dx

7 Stationary Points At some points, a curve may be neither increasing nor decreasing – we say that the curve is stationary at these points. This means that the gradient of the tangent to the curve is zero at stationary dy points, so we can find them by solving f′() x =0 or = 0 . dx The four possible stationary points are: Turning point Horizontal point of inflection Maximum Minimum Rising Falling y y y y

x x x x

A stationary point’s nature (type) is determined by the behaviour of the graph to its left and right. This is often done using a “nature table”.

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8 Determining the Nature of Stationary Points To illustrate the method used to find stationary points and determine their nature, we will do this for the graph of fx( ) =2 x3 − 9 x 2 + 12 x + 4 . Step 1 Differentiate the function. fx′( ) =6 x2 − 18 x + 12 Step 2 Find the stationary values by solving f′( x ) = 0 f′( x ) = 0. 6x2 − 18 x + 12 = 0 6()x2 − 3 x + 2 = 0(6) ÷ ()()x−1 x − 2 = 0 x=1 or x = 2 Step 3

Find the y-coordinates of the stationary f (1) = 9 so (1,9 ) is a stat. pt.

points. f (2) = 8 so (2, 8 ) is a stat. pt. Step 4 Write t he stationary values in the top row x →1 →→ 2 → of the nature table, with arrows leading in f′() x and out of them. Graph Step 5 Calculate f′( x ) for the values in the table, x →1 →→ 2 → and record the results. This gives the f′() x 0 0 gradient at these x values, so zeros confirm Graph that stationary points exist here. Step 6 → →→ → Calculate f′( x ) for values slightly lower x 1 2 and higher than the stationary values and f′() x +0 −− 0 + record the sign in the second row, e.g.: Graph f ′(0.8) > 0 so enter + in the first cell. Step 7 We can now sketch the graph near the x →1 →→ 2 → stationary points: f′() x +0 −− 0 + + means the graph is increasing and Graph − means the graph is decreasing. Step 8

The nature of the stationary points can (1,9 ) is a max. turning point.

then be concluded from the sketch. (2, 8 ) is a min. turning point.

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EXAMPLES 1. A curve has equation yx=3 −6 x 2 + 9 x − 4 . Find the stationary points on the curve and determine their nature. Given yx=3 − 6 x 2 + 9 x − 4, dy =3x2 − 12 x + 9. dx dy Stationary points exist where = 0: dx 3x2 − 12 x + 90 = 3()x2 − 4 x + 3 = 0(3) ÷ x2 −4 x + 3 = 0 ()()x−1 x − 3 = 0 x−=1 0 or x −= 3 0 x=1 x = 3. When x =1, When x = 3, y =(1)3 − 61( ) 2 + 91( ) − 4 y =(3)3 − 63( ) 2 + 93( ) − 4 =1 − 6 + 9 − 4 =27 − 54 + 27 − 4 = 0. = − 4. Therefore the point is (1,0 ) . Therefore the point is (3,− 4 ) . Nature: x →1 →→ 3 → dy + −− + dx 0 0 Graph

So (1,0 ) is a maximum turning point,

(3, − 4 ) is a minimum turning point.

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2. Find the stationary points of y=4 x3 − 2 x 4 and determine their nature. Given y= 4 x3 − 2 x 4 , dy =12x2 − 8 x 3 . dx dy Stationary points exist where = 0: dx 12x2− 8 x 3 = 0 4x2 () 32− x = 0 4x2 = 0 or 32 − x = 0 = = 3 x0 x 2 . When x = 0, 3 When x = 2 , y =40()()3 − 20 4 3 4 y =4( 3) − 2 ( 3 ) = 0. 2 2 =27 − 81 Therefore the point is (0, 0 ). 2 8 = 27 8 . 3 27 Therefore the point is ( 2, 8 ) . Nature: → →→3 → x 0 2 dy + ++ − dx 0 0 Graph

So (0, 0 ) is a rising point of inflection, 3 27 ( 2, 8 ) is a maximum turning point.

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1 3. A curve has equation y=2 x + for x ≠ 0. Find the x-coordinates of the x stationary points on the curve and determine their nature. Given y= 2 x + x −1 , dy =2 − x −2 dx 1 =2 − . x 2 dy Stationary points exist where = 0: dx 1 2− = 0 x 2 2x 2 = 1 2 1 x = 2 = ± 1 . x 2 Nature: →−1 →→ 1 → x 2 2 dy +0 −− 0 + dx Graph

So the point where x = − 1 is a maximum turning point and the point 2 where x = 1 is a minimum turning point. 2

9 Curve Sketching In order to sketch a curve, we first need to find the following:
x-axis intercepts (roots) – solve y = 0;
y-axis intercept – find y for x = 0;
stationary points and their nature.

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EXAMPLE Sketch the curve with equation y=2 x3 − 3 x 2 . y-axis intercept , i.e. x = 0: x-axis intercepts i.e. y = 0: 3 2 y =20()() − 30 2x3− 3 x 2 = 0 = 0. 2 x()2 x − 3 = 0

( ) 2 Therefore the point is 0, 0 . x = 0 or 2x − 3 = 0 x = 0 = 3 x 2 ()0, 0 3 ()2 , 0 . Given y= 2 x3 − 3 x 2 , dy =6x2 − 6 x . dx dy Stationary points exist where = 0: dx 6x2 − 6 x = 0 6x() x − 1 = 0 6x = 0 or x −1 = 0 x = 0 x =1. When x = 0, When x =1, y =20( )3 − 30( ) 2 y =21( )3 − 31( ) 2 = 0. =2 − 3

Therefore the point is (0, 0 ). = − 1. Therefore the point is (1,− 1 ) . Nature: x →0 →→ 1 → (0, 0 ) is a maximum turning point. dy + −− + dx 0 0 (1,− 1 ) is a minimum turning point. Graph y y=2 x3 − 3 x 2

O x 3 2

(1, − 1 )

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10 Closed Intervals Sometimes it is necessary to restrict the part of the graph we are looking at using a closed interval (also called a restricted domain). The maximum and minimum values of a function can either be at its stationary points or at the end points of a closed interval. Below is a sketch of a curve with the closed interval −2 ≤x ≤ 6 shaded. y

maximum

O x minimum –2 6 Notice that the minimum value occurs at one of the end points in this example. It is important to check for this whenever we are dealing with a closed interval.

EXAMPLE A function f is defined for −1 ≤x ≤ 4 by fx( ) =2 x3 − 5 x 2 − 41 x + . Find the maximum and minimum value of f( x ). Given fx( ) = 2 x3 − 5 x 2 − 4 x + 1, fx′() =6 x2 − 10 x − 4. Stationary points exist where f′( x ) = 0: 6x2 − 10 x − 40 = 23()x2 − 5 x − 2 = 0 ()()x−23 x + 1 = 0 x−=20 or 3 x += 10 = = − 1 x2 x 3 .

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To find coordinates of stationary points: 3 2 3 2 −=−1 1 −− 1 −−+ 1 f (2) = 22( ) − 52( ) − 42( ) + 1 f ( 3) 2( 3) 5( 3) 4( 3 ) 1 =16 − 20 −+ 8 1 =−−1 1 − 1 + 2()27 5() 9 41() 3 = − 11. =−2 −5 + 4 + 1 Therefore the point is (2,− 11 ). 27 9 3 = 46 27 . 1 46 − Therefore the point is ( 3, 27 ). Nature: 1 1 46 →− →→ → − x 3 2 ( 3, 27 ) is a max. turning point. ′() + −− + f x 0 0 (2,− 11 ) is a min. turning point. Graph

Points at extremities of closed interval: 3 2 f (−=−121) ( )3 −− 51( ) 2 −−+ 411( ) f (4) = 24( ) − 54( ) − 44( ) + 1 =−2 − 5 + 4 + 1 =128 − 80 − 16 + 1 = − 2. = 33.

Therefore the point is (−1, − 2 ). Therefore the point is (4, 33 ) .

Now we can make a sketch: Note y A sketch may help you

(4, 33 ) to decide on the correct answer, but it is not required in the exam. 1 46 − ( 3, 27 )

O x

(−1, − 2 ) (2, − 11 )

The maximum value is 33 which occurs when x = 4 . The minimum value is −11 which occurs when x = 2 .

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11 Graphs of Derivatives − The derivative of an xn term is an xn 1 term – the power lowers by one. For example, the derivative of a cubic (where x3 is the highest power of x) is a quadratic (where x2 is the highest power of x). When drawing a derived graph:

All stationary points of the original curve become roots (i.e. lie on the x- axis) on the graph of the derivative.

Wherever the curve is strictly decreasing, the derivative is negative. So the graph of the derivative will lie below the x-axis – it will take negative values.

Wherever the curve is strictly increasing, the derivative is positive. So the graph of the derivative will lie above the x-axis – it will take positive values. Quadratic Cubic Quartic y y y

dec. inc. dec. inc. inc. inc. dec. inc. dec. O O x O x x

y y y

+ + + + + O O x x − − O x − − Linear Quadratic Cubic

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EXAMPLE The curve y= f( x ) shown below is a cubic. It has stationary points where x =1 and x = 4 . y

y= f( x )

O x

Sketch the graph of y= f′( x ).

Since y= f( x ) has stationary points at x =1 and x = 4 , the graph of y= f′( x ) crosses the x-axis at x =1 and x = 4 . y y= f′( x ) Note The curve is increasing between the stationary O x points so the derivative is 1 4 positive there.

12 Optimisation In the section on closed intervals, we saw that it is possible to find maximum and minimum values of a function. This is often useful in applications; for example a company may have a function P( x ) which predicts the profit if £ x is spent on raw materials – the management would be very interested in finding the value of x which gave the maximum value of P( x ). The process of finding these optimal values is called optimisation . Sometimes you will have to find the appropriate function before you can start optimisation.

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EXAMPLE 1. Small plastic trays, with open tops and square bases, are being designed. They must have a volume of 108 cubic centimetres.

h x The internal length of one side of the base is x centimetres, and the internal height of the tray is h centimetres. (a) Show that the total internal surface area A of one tray is given by 432 A= x 2 + . x (b) Find the dimensions of the tray using the least amount of plastic. (a) Volume= area of base × height Let A be the surface area for a = x2 h . particular value of x : 2 We are told that the volume A= x + 4 xh . is 108 cm3, so: 108 We have h = , so: Volume= 108 x 2 x2 h =108 A= x2 + 4 x 108 ( x 2 ) 108 = 2 432 h 2 . =x + . x x (b) The smallest amount of plastic is used when the surface area is minimised. dA 432 =2x − . dx x 2 dA Stationary points occur when = 0 : Nature: dx x →6 → 432 2x − = 0 2 dA − + x dx 0 x 3 = 216 Graph x = 6. So the minimum surface area occurs when x = 6 . For this value of x : 108 h = = 3. 62 So a length and depth of 6 cm and a height of 3 cm uses the least amount of plastic.

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Optimisation with closed intervals In practical situations, there may be bounds on the values we can use. For example, the company from before might only have £100 000 available to spend on raw materials. We would need to take this into account when optimising. Recall from the section on Closed Intervals that the maximum and minimum values of a function can occur at turning points or the endpoints of a closed interval.

2. The point P lies on the graph of fx( ) = x2 −12 x + 45 , between x = 0 and x = 7 . y

y= f( x )

P( p , f( p ))

− O x p 7 A triangle is formed with vertices at the origin, P and (− p,0 ) . (a) Show that the area, A square units, of this triangle is given by 1 3 2 45 Ap=2 −6 p + 2 p . (b) Find the greatest possible value of A and the corresponding value of p for which it occurs. (a) The area of the triangle is 1 A =2 ×base × height 1 =2 ×p × f() p

1 2 =2 p() p −12 p + 45 1 3 2 45 =2p −6 p + 2 p .

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(b) The greatest value occurs at a stationary point or an endpoint. dA At stationary points = 0 : dp dA =3 p2 −12 p +=45 0 dp 2 2 3p2 − 24 p + 45 = 0 p2 −8 p + 15 = 0 ()()p−3 p − 5 = 0 p=3 or p = 5.

Now evaluate A at the stationary points and endpoints: • when p = 0, A = 0; • 1 3 2 45 when p = 3, A =×23 −× 6 3 + 2 ×= 3 27 ; • 1 3 2 45 when p =5, A =×25 −× 6 5 + 2 ×= 5 25 ; • 1 3 2 45 when p = 7 , A =×27 −× 6 7 + 2 ×= 7 35 . So the greatest possible value of A is 35, which occurs when p = 7 .

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