5 Steps to a 5 AP Calculus AB 2019 - William Ma

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MA 3972-MA-Book April 11, 2018 17:21

142 STEP 4. Review the Knowledge You Need to Score High


no maximum value. The minimum value occurs at the endpointx=3 and the minimum
value is

1


9


. (See Figure 8.1-8.)


[–1, 4] by [–1, 6]
Figure 8.1-8

8.2 Determining the Behavior of Functions


Main Concepts:Test for Increasing and Decreasing Functions, First Derivative Test and
Second Derivative Test for Relative Extrema, Test for Concavity and
Points of Inflection

Test for Increasing and Decreasing Functions
Letf be a continuous function on the closed interval [a,b] and differentiable on the open
interval (a,b).


  1. Iff′(x)>0on(a,b), thenf is increasing on [a,b].

  2. Iff′(x)<0on(a,b), thenf is decreasing on [a,b].

  3. Iff′(x) =0on(a,b), then f is constant on [a,b].


Definition:Let f be a function defined at a numberc. Thencis a critical number off if
eitherf′(c)=0orf′(c) does not exist. (See Figure 8.2-1.)

f(x)

f′ < 0
f decreasing

f′ < 0
f′ > 0 f decreasing^
f increasing

f′ = 0

f′ = 0
constant

y

0 x

f′ = 0

Figure 8.2-1
Example 1
Find the critical numbers off(x)= 4 x^3 + 2 x^2.
To find the critical numbers of f(x), you have to determine where f′(x)=0 and where
f′(x) does not exist. Notef′(x)= 12 x^2 + 4 x, and f′(x) is defined for all real numbers. Let
f′(x)=0 and thus 12x^2 + 4 x=0, which implies 4x(3x+1)= 0 ⇒x=− 1 /3orx=0.
Therefore, the critical numbers off are 0 and− 1 /3. (See Figure 8.2-2.)
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