What is the critical f(x)=4x^3+3x^2-6x+1 ?

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

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critical f(x)=4x^3+3x^2-6x+1

critical points

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

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

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

P (x, y) = (x^{2} - y^{2}) (x^{2} - 2 x y + y^{2})

y^{'} (x) + x (y) y

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