MATH 140ASpring Semester 2002Exam IIIApril 1, 2002ANSWERS:1. B; 2. A; 3. D; 4. E; 5. A; 6. D; 7. E; 8. C.9. A critical number is (the x-coordinate of) an interior point in the domain of a function where thederivative is either zero or is undefined (i.e. where f0(x) = 0 or where f0(x) does not exist).10.a. (−∞, −3)S(−3, 3)S(3, ∞)b. x = 0 (x = 3 and x = −3 are NOT critical numbers since they are not in the domain; in fact, they arethe vertical asymptotes.)c. y = 0 (i.e., the x-axis)d. x = 3 and x = −3e. (−∞, −3)S(−3, 0)f. (0, 3)S(3, ∞)g. 0h. There is none.i. (−∞, −3)S(3, ∞)j. (−3, 3)k. There is none.11. h is always negative, always increasing, and always concave down.12.f = ax3+ bx2+ cx + df0= 3ax2+ 2bx + cf00= 6ax + 2bWhere a cannot be zero, else this is not a third degree polynomial.a. No critical numbers means that f0(x) must have two complex roots, so its discriminant (2b)2−4(3a)(c)must be negative. Take, for example, a = c = 1 and b = 0 (d can be anything, so set it to 0):f(x) = x3+ xb. Two critical numbers means that f0(x) must have two distinct real roots, so its discriminant (2b)2−4(3a)(c) must be positive. Take, for example, a = 1, b = 1 and c = 0 (d can be anything, so set it to 0):f(x) = x3+ x2c. The second derivative f00must be zero at x = 3, so (x − 3) must be a factor of f00= 6ax + 2b. That’sis 2b has to be -3 times 6a, or b = −9a. f00also must be negative when x < 3 and positive when x > 3. Withthat in mind, set a = −1 and b = 9 in f (c and d can be anything, so set them to 0):f(x) = −x3+ 9x2(f(x) = x3− 9x2would have the wrong concavity.)d. The first derivatives must be zero at x = 0 and x = 2, so x and (x− 2) must both be factors of f0. Sincef0is already a quadratic polynomial, it must be a constant multiple of x(x− 2). That is, f0(x) = k(x2− 2x) =kx2− 2kx, for some nonzero number k (can let it to be 1, for now). Equate this with f0= 3ax2+ 2bx + cwe see that c must be zero and 2b is -2 times 3a, or b is -3 times a. Setting a = 1, b = −3 and c = 0, we getf(x) = kx3− 3kx2. But, checking the signs and we see that k must be a negative number, so set k = −1(again, d can be anything, so set it to 0):f(x) = −x3+ 3x2(f(x) = x3− 3x2would have a maximum at x = 0 and a minimum at x = 2.)13. The desired radius is r =12V1314. The largest possible area is 4.Page