Barrons AP Calculus - David Bock

FIGURE N6–9

† Note that the trapezoid AT 1 T 2 B is different from the trapezoids in Figure N6–7, which are like the ones we use in applying the
trapezoid rule.

SOLUTION: Since f increases on [a,b] and is concave up, the inequality is

Graphing a Function from Its Derivative; Another Look

EXAMPLE 28

Figure N6–10 is the graph of function f ′(x); it consists of two line segments and a semicircle. If f

(0) = 1, sketch the graph of f (x). Identify any critical or inflection points of f and give their

coordinates.

FIGURE N6–10

SOLUTION: We know that if f ′ > 0 on an interval then f increases on the interval, while if f ′ <

0 then f decreases; also, if f ′ is increasing on an interval then the graph of f is concave up on the

interval, while if f ′ is decreasing then the graph of f is concave down. These statements lead to

the following conclusions:

f increases on [0,1] and [3,5], because f ′ > 0 there;

but f decreases on [1,3], because f ′ < 0 there;

also the graph of f is concave down on [0,2], because f ′ is decreasing;

but the graph of f is concave up on [2,5], because f ′ is increasing.

Additionally, Since f ′(1) = f ′(3) = 0, f has critical points at x = 1 and x = 3. As x passes through