What is the critical f(x)=(x+4)(x-1)^2 ?

The critical f(x)=(x+4)(x-1)^2 is x=-7/3 ,x=1

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critical f(x)=(x+4)(x-1)^2

critical points

f (x) = (x + 4) (x - 1)^{2}

x = - \frac{7}{3}, x = 1

Solution steps

(x + 4) (x - 1)^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{7}{3}, x = 1

Identify critical points not in

f (x)

domain

Domain of

(x + 4) (x - 1)^{2} : - \infty < x < \infty

All critical points are in domain

x = - \frac{7}{3}, x = 1

f (x) = x + 5

f (x) = x 9 - x

\frac{x ^{2} + 25}{x}

y = - 3

3 - 2 - x

What is the critical f(x)=(x+4)(x-1)^2 ?
The critical f(x)=(x+4)(x-1)^2 is x=-7/3 ,x=1