# solution 7

Page 235, #16. If f(x) = x3 - 12x + 17, then f'(x) = 3x2 -12 so f'(x) = 0 when x = 2 or -2. This is a polynomial so it is continuous. Since f'(0) = -12 < 0, the derivative is negative for -2 < x < 2 and the function is decreasing on the open interval (-2,2). Since f(3) = 6 > 0, the derivative f'(x) is positive for all x > 2, and since f'(-x) = f'(x) for all x, we can conclude that f'(x) is also positive for all x , -2. The function therefore increases until x = -2, then decreases until x = 2, then increases for x > 2.

Note that the original function is odd since f(-x) = -f(x) for all x. The derivative function f'(x) on the other hand is even, since f'(-x) = f'(x) for all x. This phenomenon is true for polynomial functions in general since the derivative of a polynomial with all exponents odd will be a polynomial with all exponents even. Even more generally, if f(x) = -f(-x) then f'(x) = -f'(-x)(-x)' = f'(-x) by the chain rule, so if f is odd, f' will be even. The converse is also true, but almost the same argument.

#19. If f(x) = 3x4 + 4x3 -12x2, then f'(x) = 12x3 + 12x2 -24x = 12x(x2 + x - 2) = 12x (x+2)(x-1) = 0 when x = -2, 0, and 1. Since f'(-3) = -144 < 0, the function is decreasing for x < -2. Since f'(-1) = 24 > 0, the function increases from 0 to 2. Since f'(1/2) = -15/2 < 0, the function decreases from 0 to 1 and since f'(2) = 96 > 0, the function increases for x larger than 1. Note that f(-2) will have to be relative minimum and x = 1 will also be a relative minimum. Since f(-2) = -32 and f(1) = -5, the absolute minimum is at f = -2. Since the term with the highest exponent is even and since the coefficient of this term is positive, the function grows without bound as x gets further and further out on the positive x axis or on the negative x-axis. There is therefore no absolute maximum.

#24. If f(x) = ln(2x)/x for x > 0, then f'(x) = [x(ln(2x))' - ln(2x)(x')]/x2 = [x(1/2x)(2x)' -ln(2x)]/x2 = [1-ln(2x)]/x2. Thus f'(x) = 0 if 1 = ln(2x) so e = 2x and x = e/2. Since f'(1/2) = [1 - ln(1)]/(1/2)2 = 4 > 0, the function increases from 0 until e/2. For x > e/2, we have ln(2x) > ln(e) = 1 so the numerator of f'(x) is negative and the function is decreasing for x > e/2. This means that x = e/2 is an absolute maximum for the function over the positive real axis where x > 0.

#36. If f(x) = x + 1/x, then f is defined for all x not equal to 0, and f is defined and differentiable in the closed interval [2,3] since this interval does not contain 0 and since the derivative f'(x) = 1 - 1/x2. The average slope is [f(3)-f(2)]/[3-2] = (3 + 1/3)-(2 + 1/2) = 5/6 and this equals f'(x) when 5/6 = 1 - 1/x2 or x2 = 6. Thus x is the positive square root of 6, which is between 2 and 3 as predicted by the Mean Value Theorem.

#46. Let v(t) denote the velocity of the car at time t, so v'(t) denotes the acceleration at time t. Then [v(2:35)-v(2:25)]/[2:35-2:25] = [65 - 50]/(1/6) since the ten-minute interval is 1/6 of an hour. Therefore he average velocity over this period is 15x6 = 90 so by the Mean Value Theorem, there had to be a point between 2:25 and 2;35 when the acceleration v'(t) was exactly 90m/h2. Page 244, #12. If f(x) = x2 + 1/x, then f'(x) = 2x -1/x2 so f'(x) = 0 if x3 = 1/2. The function is undefined if x = 0, so the domain breaks up into two parts, x < 0 and x > 0. The derivative is negative for all x < 0 so the function is decreasing there. If x3 > 1/2 then f'(x) > 0 and if 0 < x3 <1/2, then f'(x) < 0. It follows that x = (1/2)1/3 gives a global minimum for the function f over the portion of the domain where x is positive, but this is not a global minimum for the entire function since f(-1) = -1 < 0, lover than the (positive) value of f at the only critical point. There are no local maximum and there are no global maxima or minima for the function.

#14. If f(x) = x2e-x/3, then f'(x) = (x2)'e-x/3 + x2(e-x/3)' = 2xe-x/3+(-1/3)x2e-x/3 = x(2-(1/3)x)e-x/3 = 0 when x = 0 or x = 6. For x < 0, f'(x) < 0 so the function is decreasing for x < 0. From 0 to 6 the funcion is increasing and after 6 it is decreasing. It follows that x = 0 is a relative minimum and x = 6 is a relative maximum.

#30. If a box has width x and length 2x and height h, then the volume is V = x(2x)h. Since this volume is fixed at 576, we have 576 = 2x2h and h = 288/x2. The total surface area of the box is S = 2xh + 2(2x)h + 2x(2x) = 6xh + 4x2 = 6x(288/x2) + 4x2 = 1728/x + 4x2. This expresses S(x) as a function of x alone, so we may find the critical points of S(x) by computing S'(x) and setting this value equal to zero. We find S'(x) = -1728/x2 + 8x = 0 if 8x3 = 1728 so x is the cube root of 216, i.e. 6. The sides of the box are then 6, 12, and 288/36 = 8. The function S(x) is undefined when x = 0 and as x approaches 0, the value of the surface area grows without bound. It follows that the critical point x = 6 must be a minimum. (In the next week of lectures we will find an effective way of determining explicitly that the critical point is a minimum. For now we may note that S/(x) > 0 if x > 6 and S'(x) < 0 if 0 < x < 6, establishing the fact that x = 6 is at least a local minimum.

#39. The cost of a can with radius r and height h will be 2(4)(pi)2 + 2(2)(pi)h while the volume V = (pi)2h is fixed at 100 so h = 100/[(pi)r2]. The cost is thus a function of x given by C(x) = 8r2 + 400/r. The derivative is C'(x) = 16(pi)r - 400/r2 and this will equal zero when r3 = 25/(pi).

#49. The volume of the floored tent is V = (4/3)x2h so h(x) = 3V/(4x2). The surface area is S(x) = 4x2 + 4(1/2)(2x)(x2 + h2)1/2. The derivative of S(x) is then S'(x) = 8x + 4(x2 + h2)1/2 + 4x(1/2)(x2 + h2)-1/2(2x + 2h(x)h'(x)).
Plugging in the value for h(x) and the derivative h'(x) = -3V/(2x3, we get S'(x) = [8x(x2 + h2)1/2 + 8x2 -9/4V2/x4]/(x2 + h2)1/2 after some simplification). Setting the numerator of this expression equal to zero, we get 8x(x2 + h2)1/2 + 8x2 -9/4V2/x4 = 0 so -8x(x2 + h2)1/2 = 8x2 -9/4V2/x4. Squaring both sides and cancelling the term 64x4 leads to the expression 72V2 = (81/16)V4/x8 so x6 = (9/128)V2 and finally x = 1/2((9/2)V2)1/6. Note that the answer in the back of the book is for the entire base, i.e. 2x rather than x.
There is no maximum value for the surface area, and this unique critical point will be the minimum value for the surface area.

Page 256, #24. f(x) = (x-2)2(2x+3)2
f'(x) = 2(x-2)(2x+3)2 +2(x-2)2(2x+3)2 = 2(x-2)(2x+3)[2x+3 +2(x-2)] = 2(x-2)(2x+3)[4x-1] = 0 if x = 2 or x = -3/2 or x = 1/4. The function is always non-negative. Since f(-3/2) = 0 and f(2) = 0, these values will both be absolute minima for f. Since f(1/4) > 0, the function must be increasing from x = -3/2 to x = 1/4, then decreasing back to zero at x = 2. It follows that x = 1/4 is a relative maximum. For x = 3, the derivative is positive so for all x > 2 the derivative is positive and the function is increasing. For x = -2, f'(x) is negative so f'(x) is negative for all x <-3/2. The graph resembles a curvy "W" resting on the x-axis.

#40 f(x)= 3x5 - 25x3 + 60x.
f'(x) = 15x4 -75x2 + 60 = 15(x2-4)(x2-1) = 0 is x = -2, -1, 1, and 2. Note that f'(0) > 0 so the function increases from x = -1 ro x = 1. Since f(1) > 0 and f(2) < 0, the function decreases from x = 1 to x = 2 and since f'(3) > 0, the function increases for x > 2. Since f(-x) = -f(x), the function is symmetric with respect to the origin, so it increases for x < -2 and decreases from x = -2 to x = -1. The shape of the graph is similar to that of Figure 4.5.20.

#48. f(x) = x1/3(2-x)2/3.
f'(x) = (1/3)x-2/3(2-x)2/3 + (2/3)(-1)x1/3(2-x)-1/3 =x-2/3(2-x)-1/3[(1/3)(2-x) - (2/3)x] = 0 if 2/3 - x = 0 so x = 2/3 is the only critical point. The function increases up to x = 2/3 and then decreases. The tangent line is vertical at x = 0. At x = 2, the graph has a cusp.

#52. We are given f(-3) = -130, f(0) = 5, and f(1) = -2. The picture of the derivative looks like a cubic with roots at -3, 0. and 2. The derivative is positive between -3 and 0 and after 1 so the function increases from x = -3 to 0 and after 1. It decreases otherwise. The graph of the function is similar to Figure 4.5.18.

Thomas Banchoff