Math 2250 HW #10 Solutions

Math 2250 Written HW #10 Solutions

1. Find the absolute maximum and minimum values of the function g(x) = e−x2 subject to the constraint −2 ≤ x ≤ 1. Answer: First, we check for critical points of g, which is diﬀerentiable everywhere. By the Chain Rule, 2 2 g0(x) = e−x · (−2x) = −2xe−x . Since e−x2 > 0 for all x, g0(x) = 0 when 2x = 0, meaning when x = 0. Hence, x = 0 is the only critical point of g. Now we just evaluate g at the critical point and the endpoints:

2 g(−2) = e−(−2) = e−4 = 1/e4 2 g(0) = e−0 = e0 = 1 2 g(2) = e−1 = e−1 = 1/e.

Since 1 > 1/e > 1/e4, we see that the absolute maximum of g(x) on this interval is at (0, 1) and the absolute minimum is at (−2, 1/e4).

2. Find all local maxima and minima of the curve y = x2 ln x. Answer: Notice, first of all, that ln x is only defined for x > 0, so the function x2 ln x is likewise only defined for x > 0. This function is differentiable on its entire domain, so we differentiate in search of critical points. Using the Product Rule, 1 y0 = 2x ln x + x2 · = 2x ln x + x = x(2 ln x + 1). x Since we’re only allowed to consider x > 0, we see that the derivative is zero only when 2 ln x + 1 = 0, meaning ln x = −1/2. Therefore, we have a critical point at x = e−1/2. For x > 0, the expression x(2 ln x + 1) has the same sign as 2 ln x + 1, which is negative for x < e−1/2 and positive otherwise. Therefore, the curve has a local minimum when x = e−1/2, √1 1 meaning at the point ( e , − 2e ). There are no other local extrema of this curve. 3. A general cubic function has the form

f(x) = ax3 + bx2 + cx + d

where a, b, c, and d are constants.

(a) Give examples that demonstrate such functions can have 0, 1, or 2 critical points. Answer: Suppose f(x) = x3 + x. Then f 0(x) = 3x2 + 1, which is always positive, so this is an example of a cubic function with no critical points. If f(x) = x3, then f 0(x) = 3x2, which is zero only when x = 0, so this is a cubic function with exactly 1 critical point. Finally, if f(x) = x3 + x2 − x + 1, then f 0(x) = 3x2 + 2x − 1 = (3x − 1)(x + 1), so this cubic function has critical points when x = 1/3 and x = −1. Thus, this cubic function has exactly two critical points.

1 (b) Show that no cubic function can have more than 2 critical points. Answer: For any cubic function f(x) = ax3 + bx2 + cx + d, we know that

f 0(x) = 3ax2 + 2bx + c.

Therefore, f 0(x) = 0 if and only if √ −2b ± p(2b)2 − 4(3a)(c) −2b ± 4b2 − 12ac x = = . 2(3a) 6a

There are at most such two values of x, so we see that f has at most two critical points. (c) How many local extreme values (maxima and/or minima) can a cubic function have? Answer: A cubic function can certainly have no local extreme values, as in the case of f(x) = x3 + x, which we saw in part (a) has no critical points. A cubic function can also have two local extreme values (1 max and 1 min), as in the case of f(x) = x3 + x2 + x + 1, which has a local maximum at x = −1 and a local minimum at x = 1/3. These are the only options. Since a cubic function can’t have more than two critical points, it certainly can’t have more than two extreme values. Also, a cubic function cannot have just one local extremum except in the slightly dumb case when a = 0 (in which case it’s really a quadratic function instead of a cubic).