What is the critical f(x)=x^3-3/2 x^2 ?

The critical f(x)=x^3-3/2 x^2 is x=0,x=1

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critical f(x)=x^3-3/2 x^2

critical points

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

x = 0, x = 1

Solution steps

x^{3} - \frac{3}{2} x^{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

x^{3} - \frac{3}{2} x^{2} : - \infty < x < \infty

All critical points are in domain

x = 0, x = 1

f (x) = 8 x^{3} - 6 x + 4

4 - x^{2}

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

y = \frac{x}{x ^{2} + 1}

(38.7, 38.24), (84.6, 87.04)

What is the critical f(x)=x^3-3/2 x^2 ?
The critical f(x)=x^3-3/2 x^2 is x=0,x=1