FIGURE N4–4
Verify the graph and information obtained above on your graphing calculator.EXAMPLE 13Sketch 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 relative
minimum. Since f ′′(0) = 0, the second-derivative test fails to tell us whether x = 0 yields a
maximum or a minimum.
(4) Since f ′(x) does not change sign as x increases through 0, the point (0, 0) yields neither a
maximum nor a minimum.
(5) f ′′(x) = 0 when x is 0 or 2; f ′′ changes signs as x increases through 0 (+ to −), and also as x
increases through 2 (− to +). Thus both (0, 0) and (2, −16) are inflection points of the curve.
The curve is sketched in Figure N4–5.
FIGURE N4–5
Verify the preceding on your calculator.