What is the critical f(x)=8x^3-12x^2+4 ?

The critical f(x)=8x^3-12x^2+4 is x=0,x=1

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critical f(x)=8x^3-12x^2+4

critical points

f (x) = 8 x^{3} - 12 x^{2} + 4

x = 0, x = 1

Solution steps

8 x^{3} - 12 x^{2} + 4

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

8 x^{3} - 12 x^{2} + 4 : - \infty < x < \infty

All critical points are in domain

x = 0, x = 1

(5 - x)^{4}

(1, 5), (2, 7)

\frac{x - 2}{3 x + 5}

\frac{4}{x} + x^{4}

- 5 sin (15 π t)

What is the critical f(x)=8x^3-12x^2+4 ?
The critical f(x)=8x^3-12x^2+4 is x=0,x=1