What is the critical 4x^3-240x^2+3.6 ?

The critical 4x^3-240x^2+3.6 is x=0,x=40

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critical 4x^3-240x^2+3.6

critical points

4 x^{3} - 240 x^{2} + 3.6

x = 0, x = 40

Solution steps

4 x^{3} - 240 x^{2} + 3.6

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 40

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} - 240 x^{2} + 3.6 : - \infty < x < \infty

All critical points are in domain

x = 0, x = 40

(2) (10)

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

\frac{3}{4} x^{5} + 5

\frac{x ^{7}}{7}

(1, 2), (22, 3)