What is the critical f(x)=2x^3+3x^2-36x+4 ?

The critical f(x)=2x^3+3x^2-36x+4 is x=-3,x=2

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critical f(x)=2x^3+3x^2-36x+4

critical points

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 4

x = - 3, x = 2

Solution steps

2 x^{3} + 3 x^{2} - 36 x + 4

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 2

Identify critical points not in

f (x)

domain

Domain of

2 x^{3} + 3 x^{2} - 36 x + 4 : - \infty < x < \infty

All critical points are in domain

x = - 3, x = 2

3 x^{4} - 4 x^{3} - 12 x + 1

f (x) = \frac{( 2 x - 1 )}{x ^{2} - 4 x - 5}

F (t) = 2 t \frac{2}{3} + t \frac{5}{3}

f (x) = 7 (x - 2)^{\frac{2}{3}}

\frac{d}{d x} (\frac{2 x y}{x ^{2} + y ^{2}})

What is the critical f(x)=2x^3+3x^2-36x+4 ?
The critical f(x)=2x^3+3x^2-36x+4 is x=-3,x=2