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

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

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critical f(x)=(x+1)^2(x-4)^3

critical points

f (x) = (x + 1)^{2} (x - 4)^{3}

x = - 1, x = 1, x = 4

Solution steps

(x + 1)^{2} (x - 4)^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 1, x = 4

Identify critical points not in

f (x)

domain

Domain of

(x + 1)^{2} (x - 4)^{3} : - \infty < x < \infty

All critical points are in domain

x = - 1, x = 1, x = 4

f (x) = 8 x^{4} - 48 x^{2}

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

r (x) = \frac{6 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 6 x}

(3, - 2), (- 4, 5)

f (x) = 5 x^{2} + 9

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