19.
20.
Part B
The graph shown below satisfies all five conditions. So do many others!
(a) f ′ is defined for all x in the interval. Since f is therefore differentiable,
it must also be continuous.
(b) Because f ′(−2) = 0 and f ′ changes from negative to positive there, f
has a local minimum at x = −2. To the left of x = 9, f ′ is negative, so f is
decreasing as it approaches that endpoint and reaches another local
minimum there.
(c) Because f ′ is negative to the right of x = −3, f decreases from its left
endpoint, indicating a local max there. Also, f ′(7) = 0 and f ′ changes
from positive to negative there, so f has a local relative maximum at x =
7.
(d) Note that . Since there is more area above the x-
axis than below the x-axis on [−3,7], the integral is positive and f(7) −
f(−3) > 0. This implies that f(7) > f(−3), and that the absolute maximum
occurs at x = 7.
(e) At x = 2 and also at x = 6, f ′ changes from increasing to decreasing,
indicating that f changes from concave upward to concave downward at
each. At x = 4, f ′ changes from decreasing to increasing, indicating that f
