Tutorial 8: Solutions

Applications of the Derivative 1. We are asked to ﬁnd the absolute maximum and minimum values of f on the given interval, and state where those values occur:

(a) f(x) = 2x3 + 3x2 12x on [1, 4]. − The absolute extrema occur either at a critical point (stationary point or point of non-diﬀerentiability) or at the endpoints. The stationary points are given by f 0(x) = 0, which in this case gives

6x2 + 6x 12 = 0 = x = 1, x = 2. − ⇒ − Checking the values at the critical points (only x = 1 is in the interval [1, 4]) and endpoints:

f(1) = 7, f(4) = 128. − Therefore the absolute maximum occurs at x = 4 and is given by f(4) = 124 and the absolute minimum occurs at x = 1 and is given by f(1) = 7. −

(b) f(x) = (x2 + x)2/3 on [ 2, 3] − As before, we ﬁnd the critical points. The derivative is given by 2 2x + 1 f 0(x) = 3 (x2 + x)1/2

and hence we have a stationary point when 2x + 1 = 0 and points of non- diﬀerentiability whenever x2 + x = 0. Solving these, we get the three critical points, x = 1/2, and x = 0, 1. − − Checking the value of the function at these critical points and the endpoints:

f( 2) = 22/3, f( 1) = 0, f( 1/2) = 2−4/3, f(0) = 0, f(3) = (12)2/3. − − − Hence the absolute minimum occurs at either x = 1 or x = 0 since in both − these cases the minimum value is 0, while the absolute maximum occurs at x = 3 and is f(3) = (12)2/3.

f 0(x) = 1 2 cos x −

1 and the only critical point on [ π/4, π/2] is x = π/3. Checking the value of − the function at this critical point and at the end points: f( π/4) = π/4 + √2 = 0.6288 − − f(π/3) = π/3 √3 = 0.6849 − − f(π/2) = π/2 2 = 0.4292. − − Hence the absolute minimum is at x = π/3 and is given by f(π/3) above while the absolute maximum is at x = π/4. −

(d) f(x) = 6 4x on [ 3, 3] | − | − The absolute value of a continuous differentiable function is everywhere differentiable except at the points where what’s inside the absolute value vanishes, i.e., the only critical point is the point of non-differentiablility given by 6 4x = 0 = x = 3/2. − ⇒ We now check the value of the function at this critical point and at the endpoints: f( 3) = 18, f(3) = 6, f(3/2) = 0. − Hence the absolute minimum occurs at x = 3/2 and is given by f(3/2) = 0 and the absolute maximum occurs at x = 3 and is given by f( 3) = 18. − −

(e) f(x) = x2 x 2 on ( , ) − − −∞ ∞ A polynomial of even degree over the entire real line will have only one global extremum which occurs at a stationary point, deﬁned by f 0(x) = 2x 1 = 0, = x = 1/2. − ⇒ It should be clear from the limiting behaviour as x that this point → ±∞ has to be an absolute minimum. Alternatively, we could compute the second derivative f 00(x) = 2 > 0, which implies a minimum. So there is an absolute minimum at x = 1/2 given by f(1/2) = 9/4. There is no absolute maximum. −

(f) f(x) = x3 9x + 1 on ( , ) − −∞ ∞ This is a polynomial of odd degree and hence has no absolute extrema. 2. A closed rectangular container with a square base is to have a volume of 2000 cm3. It costs twice as much per square centimeter for the top and bottom as it does for the sides. Find the dimensions of the container of least cost.

The volume is constrained by the equation V = x2h = 2000,

2 where x is the length of the side of the square base and h is the height. We are trying to minimize the cost so we are looking for a function which represents the total cost of manufacturing the container. Let p be the price/cm2 for the sides and hence 2p is the price/cm2 for the top and bottom of the container. We have

cost of top and bottom = (area) (price/cm2) × = (2x2) (2p) = 4x2p × cost of sides = (area) (price/cm2) × = (4xh) (p) = 4xhp. × Hence the total cost is C = 4x2p + 4xhp. This is a function of both h and x so we use the constraint equation to eliminate the h-dependence, 2000 h = x2 and hence the cost function is 8000p C(x) = 4x2p + . x The absolute extrema on an open interval will occur at the critical points. The stationary points are deﬁned by 8000p C0(x) =8xp = 0 − x2 = x3 = 1000 ⇒ = x = 10 cm. ⇒ There is also a point of non-diﬀerentiability at x = 0 but this clearly isn’t in the domain of interest. It is straighforward to show that f 00(10) = 24p > 0 and hence x = 10 cm is indeed the dimension of minimum cost. Subbing into the constraint equation gives h = 20 cm.

3. The equation f(x) = x5 + x4 5 = 0 has one real solution. Use the Intermedi- − ate Value Theorem to obtain an initial estimate of the solution, then apply the Newton-Raphson method to approximate the root to 5 decimal places.

Evaluating f(1) = 3 < 0 and f(2) = 43 > 0. Since the sign changes and − the function is continuous, the Intermediate Value Theorem tells us that there must be a root in the interval (1, 2). Taking the midpoint of this interval to be our

3 initial guess x1 = 1.5. Then the Newton-Raphson method gives

f(x1) x2 = x1 0 = 1.3027375201 − f (x1) f(x2) x3 = x2 0 = 1.232510136 − f (x2) f(x3) x4 = x3 0 = 1.22453239 − f (x3) f(x4) x5 = x4 0 = 1.224439563 − f (x4) f(x5) x6 = x5 0 = 1.22443955. − f (x5) So to 5 decimal places, we have x = 1.22444.

4. Apply Rolle’s Theorem and use the fact that d (3x4 + x2 4x) = 12x3 + 2x 4 dx − − to show that the equation 12x3 +2x 4 = 0 has at least one solution in the interval − (0, 1).

We consider the function f(x) = 3x4 + x2 4x, which is continuous and dif- − ferentiable on [0, 1] and which satisfies f(0) = 0, f(1) = 0. Hence Rolle’s Theorem 0 implies that there is at least one point x0 (0, 1) such that f (x0) = 0, i.e., x0 is ∈ a root of 12x3 + 2x 4 = 0, as required. − 5. For the function f(x) = √25 x2, verify that the hypothesis of the Mean-Value − Theorem is satisfied on the interval [ 5, 3] and find all values of c in that interval − that satisfy the conclusion of the theorem.

The hypothesis of the Mean Value Theorem requires that the function be continuous on some closed interval [a, b] and diﬀerentiable on the open interval (a, b). In the present case, f(x) is continuous on the domain consisting of values satisfying

25 x2 0, = (f) = [ 5, 5]. − ≥ ⇒ D − Computing the derivative x f 0(x) = , −√25 x2 − we see that the function is diﬀerentiable on the open interval ( 5, 5). Hence − the hypothesis of the Mean Value Theorem is satisﬁed on the interval [ 5, 3]. −

4 The coordinates of the endpoints of this interval are A( 5, f( 5)) = ( 5, 0) and − − − B(3, f(3)) = (3, 4) and the slope of this line-segent is 4 0 1 slope of AB = − = . 3 ( 5) 2 − − The Mean Value Theorem states that there is at least one value c such that f 0(c) is equal to the slope of this line segment, i.e., c 1 = −√25 c2 2 − = c2 = 5 ⇒ = c = √5. ⇒ ± 6. The function s(t) = t4 4t3 + 4t2 + 1, t 0, describes the position of a particle − ≥ moving along a coordinate line, where s is in metres and t is in seconds. Analyze the motion of the particle by determining the intervals over which the particle is moving in the positive or negative direction, the times when the particle has stopped and the intervals over which the particle is speeding up or slowing down. Give a schematic picture of the motion.

The velocity is given by

v(t) = s0(t) = 4t3 12t2 + 8t = 4t(t 1)(t 2). − − − Hence the velocity is zero at t = 0, t = 1 and t = 2 which correspond to the particle’s turning points. The sign of the velocity determines whether the particle is moving in the positive or negative s-direction: The acceleration is

______+ + + + + + + + + + + + + + + + + + + + 0 0 + + + + + + + + + + + + + + + + + + + + sign of v(t) t 0 1 2 moving in pos. dir. moving in neg. dir. moving in pos. dir.

a(t) = 12t2 24t + 8 = 4(3t2 6t + 2) − − which has roots at 1 t = 1 . ± √3 The intervals on which the sign of the acceleration and velocity are the same are the intervals over which the particle is speeding up and vice-versa.

5 ______+ + + + + + + + + 0 0 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + sign of a(t) ______+ + + + + + + + + + + + + + + + + + + + 0 0 + + + + + + + + + + + + + + + + + + + + sign of v(t)

1 1 t 1 1+ 0 − √3 1 √3 2 speeding up slowing down speeding up slowing down speeding up

Finally, to draw the schematic diagram, we note the position of the particle along the s-axis at each of these critical times:

s(0) = 1, s(1) = 2, s(2) = 1, s(1 1/√3) = 1.4444‘ ± . The schematic diagram is

1 t = 1 + t =2 √3 t =1 t =0 1 t =1 − √3 s 1 2

Figure 1: The red shaded regions are the intervals over which the particle is slowing down