Figure N4–4Verify the graph and information obtained above on your graphing calculator.
Example 13 __
Sketch the graph of f (x) = x^4 − 4x^3.
SOLUTION:
(1) f ′(x) = 4x^3 − 12x^2 and f ′′(x) = 12x^2 − 24x.
(2) f ′(x) = 4x^2 (x − 3), which is zero when x = 0 or x = 3.
(3) Since f ′′(x) = 12x(x − 2) and f ′′(3) > 0 with f ′(3) = 0, the point (3, −27) is a
local minimum. Since f ′′(0) = 0, the second-derivative test fails to tell us
whether x = 0 yields a local maximum or a local minimum.
(4) Since f ′(x) does not change sign as x increases through 0, the point (0, 0)
yields neither a local maximum nor a local minimum.
(5) f ′′(x) = 0 when x is 0 or 2; f ′′ changes signs as x increases through 0 (+ to