1.
2.
3.
4.
5.
6.
7.
8
Part A
(B) Use the Rational Function Theorem.
(B) Note that , where f(x) = ln x.
(D) Since y′ = −2xe−x
2
, therefore y′′ = −2(x · e−x
2
· (−2x) + e−x
2
). Replace
x by 0.
(B).
(B) h′(3) = g′ (f(3)) · f ′(3) = g′(4) · f ′(3) = · 2.
(B) Since f ′(x) exists for all x, it must equal 0 for any x 0 for which f is a
relative minimum, and it must also change sign from negative to positive
as x increases through x 0 . For the given derivative, this only occurs at x =
−1.
(C) By the Quotient Rule (formula (6),
(A) Here, f ′(x) is e-x (1 − x); f has maximum value when x = 1.
