What is the critical f(x)=(x+3)(x-5)^2 ?

The critical f(x)=(x+3)(x-5)^2 is x=-1/3 ,x=5

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critical f(x)=(x+3)(x-5)^2

critical points

f (x) = (x + 3) (x - 5)^{2}

x = - \frac{1}{3}, x = 5

Solution steps

(x + 3) (x - 5)^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{3}, x = 5

Identify critical points not in

f (x)

domain

Domain of

(x + 3) (x - 5)^{2} : - \infty < x < \infty

All critical points are in domain

x = - \frac{1}{3}, x = 5

f (x) = x - \frac{1}{5}

f (x) = \frac{2 x}{x + 1}

f (x) = 5 \cdot 3^{x}

x^{2} e^{1 - x^{2}}

(4, 2), (- 2, 4)

What is the critical f(x)=(x+3)(x-5)^2 ?
The critical f(x)=(x+3)(x-5)^2 is x=-1/3 ,x=5