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Kulotskie
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Tyro
Asked: 2021-04-17T10:51:19+00:00 In: ,

The uses of the first and second derivative to determine the intervals of increase and decrease of a function.

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For the function f(x) = (x + 3)(x – 2) 3 find
a) the interval(s) of increase and decrease of f,

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  2. acedstud
    This answer was edited.

    Given a function, f, by the first derivative test, the function is increasing in the intervals where f'>0 and decreasing in the intervals where f'<0.

    Thus, given f(x)=(x+3)(x-2)^3, f'(x)=(x+3)\cdot3(x-2)^2+(x-2)^3=3(x+3)(x-2)^2+(x-2)^3

    =(x-2)^2(3x+9+x-2)=(x-2)^2(4x+7).

    The turning poions are f'(x)=0\Rightarrow(x-2)^2(4x+7)=0\Rightarrow x=2 or x=-\dfrac{7}{4}.

    Thus, the turning points divives the domain into three regions, namely \left(-\infty, -\dfrac{7}{4}\right), \left(-\dfrac{7}{4}, 2\right), and (2, \infty).

    Testing points on the interval \left(-\infty, -\dfrac{7}{4}\right), f'(-2)=(-2-2)^2(4(-2)+7)=(-4)^2(-8+7)=(16)(-1)=-16<0.

    Thus, the function is decreasing on the interval \left(-\infty, -\dfrac{7}{4}\right)

    Testing points on the interval \left(-\dfrac{7}{4}, 2\right), f'(0)=(0-2)^2(4(0)+7)=(-2)^2(0+7)=(4)(7)=28>0.

    Thus, the function is increasing on the interval \left(-\dfrac{7}{4}, 2\right).

    Testing points on the interval (2, \infty), f'(3)=(3-2)^2(4(3)+7)=(1)^2(12+7)=(1)(19)=19>0.

    Thus, the function is increasing on the interval (2, \infty).

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