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

The critical 3x^4-4x^3+2 is x=0,x=1

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

critical points

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

x = 0, x = 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 0, x = 1

2 + \frac{6}{x}, x > 0, y > 2

f (x) = 2 x^{3} + 6 x^{2} - 144 x + 4

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

f (x) = \frac{x ^{2}}{x + 6}