SOLUTIONS: See Figure N6–21b.
(a) g(b) is the slope of G(x) at b, the slope of line ST.
(b) is equal to the area under G(x) from a to b.
(c) = G(b) − G(a) = length of BT − length of BR = length of RT.
(d)
EXAMPLE 43
The function f (t) is graphed in Figure N6–22a. Let
FIGURE N6–22a
(a) What is the domain of F?
(b) Find x, if F ′(x) = 0.
(c) Find x, if F(x) = 0.
(d) Find x, if F(x) = 1.
(e) Find F ′(6).
(f) Find F(6).
(g) Sketch the complete graph of F.
SOLUTIONS:
(a) The domain of f is [−2,1] and [2,6], We choose the portion of this domain that contains the
lower limit of integration, 4. Thus the domain of
F is 2 ≤ ≤ 6, or 4 ≤ x ≤ 12.
(b) Since
(c) F(x) = 0 when or x = 8. F(8) =
(d) For F(x) to equal 1, we need a region under f whose left endpoint is 4 with area equal to 1.
The region from 4 to 5 works nicely; so and x = 10.
(e)