(B) We want to approximate the change in area of the square when a side
of length e increases by 0.01e. The answer is
A′(e)(0.01e) or 2 e(0.01e).
(D) Since V = e^3 , V′ = 3e^2 . Therefore at e = 10, the slope of the tangent
line is 300. The change in volume is approximately 300(±0.1) = 30 in.^3
(E) f ′(x) = 4x^3 − 8x = 4x(x^2 − 2). f ′ = 0 if x = 0 or ±.
f ′′(x) = 12x^2 − 8; f ′′ is positive if x = ± , negative if x = 0.
(C) Since f ′′(x) = 4(3x^2 − 2), it equals 0 if . Since f ′′ changes sign
from positive to negative at and from negative to positive at
both locate inflection points.
(A) The domain of y is {x | x 2}. Note that y is negative for each x in
the domain except 2, where y = 0.
(B) f ′(x) changes sign (from negative to positive) as x passes through
zero only.
(E) The graph must be decreasing and concave downward.
