Test for Increasing and Decreasing Functions for AP Calculus
Practice problems for these concepts can be found at: Graphs of Functions and Derivatives Practice Problems for AP Calculus
Let f be a continuous function on the closed interval [a, b] and differentiable on the open interval (a, b).
- If f '(x) > 0 on (a, b), then f is increasing on [a, b].
- If f '(x) < 0 on (a, b), then f is decreasing on [a, b].
- If f '(x) = 0 on (a, b), then f is constant on [a, b].
Definition: Let f be a function defined at a number c. Then c is a critical number of f if either f '(c) = 0 or f '(c) does not exist. (See Figure 7.2-1.)
Find the critical numbers of f(x) = 4x3 + 2x2.
To find the critical numbers of f(x), you have to determine where f '(x) = 0 and where f '(x) does not exist. Note f '(x) = 12x2 + 4x, and f '(x) is defined for all real numbers. Let f '(x) = 0 and thus 12x2 + 4x = 0 which implies 4x(3x +1) = 0 x = – 1/3 or x = 0. Therefore the critical numbers of f are 0 and –1/3. (See Figure 7.2-2.)
Find the critical numbers of
Note that f '(
The graph of f ' on (1, 6) is shown in Figure 7.2-4. Find the intervals on which f is increasing or decreasing.
See Figure 7.2-5.
Thus, f is decreasing on [1, 2] and [5, 6] and increasing on [2, 5].
Find the open intervals on which is increasing or decreasing.
Step 1: Find the critical numbers of f.
Set f '(x) = 0 4x = 0 or x = 0.
Since f '(x) is a rational function, f '(x) is undefined at values where the denominator is 0. Thus, set x2 – 9 = 0 x = 3 or x = – 3. Therefore the critical numbers are –3, 0, and 3.
Step 2: Determine intervals.
Intervals are (–∞, –3), (–3, 0), (0, 3), and (3,∞).
Step 3: Set up a table.
Step 4: Write a conclusion. Therefore f(x) is increasing on [–3, 0] and [3, ∞) and decreasing on ([–∞, [–3] and [0, 3]. (See Figure 7.2-6.)
The derivative of a function f is given as f '(x) = cos(x2). Using a calculator, find the values of x on such that f is increasing. (See Figure 7.2-7.)
Using the [Zero] function of the calculator, you obtain x =1.25331 is a zero of f ' on Since f '(x)= cos(x2) is an even function, x = –1.25331 is also a zero on
(See Figure 7.2-8.)
Thus f is increasing on [–1.2533, 1.2533].
Practice problems for these concepts can be found at:
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