What is the critical 4x^3+3x^2-6x+1 ?

The critical 4x^3+3x^2-6x+1 is x=-1,x= 1/2

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critical 4x^3+3x^2-6x+1

critical points

4 x^{3} + 3 x^{2} - 6 x + 1

x = - 1, x = \frac{1}{2}

Solution steps

4 x^{3} + 3 x^{2} - 6 x + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} + 3 x^{2} - 6 x + 1 : - \infty < x < \infty

All critical points are in domain

x = - 1, x = \frac{1}{2}

f (x, y) = 2 x^{2} + 3 x y + y^{2} + a x + 5

f (x) = x^{3} - 4 x

x^{3} - 12 x + 1

sin (2 x), 0 \leq x \leq 2 π

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

What is the critical 4x^3+3x^2-6x+1 ?
The critical 4x^3+3x^2-6x+1 is x=-1,x= 1/2