What is the critical f(x)=2x^3-96x+42 ?

The critical f(x)=2x^3-96x+42 is x=-4,x=4

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critical f(x)=2x^3-96x+42

critical points

f (x) = 2 x^{3} - 96 x + 42

x = - 4, x = 4

Solution steps

2 x^{3} - 96 x + 42

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

2 x^{3} - 96 x + 42 : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 4

f (x) = \frac{( x + 2 )}{x ^{2} + 4 x - 5}

y = e^{x + 1}

f (x) = x^{2} (x - a)

y = \frac{- 2}{3} x + 13

f (x) = x - 1 - 2

What is the critical f(x)=2x^3-96x+42 ?
The critical f(x)=2x^3-96x+42 is x=-4,x=4