What is the extreme f(x)=x^3+2x^2 ?

The extreme f(x)=x^3+2x^2 is Maximum(-4/3 , 32/27),Minimum(0,0)

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extreme f(x)=x^3+2x^2

extreme points

f (x) = x^{3} + 2 x^{2}

Maximum (- \frac{4}{3}, \frac{32}{27}), Minimum (0, 0)

Solution steps

f^{'} (x) = 3 x^{2} + 4 x

Find intervals:

Increasing

: - \infty < x < - \frac{4}{3},

Decreasing

: - \frac{4}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{4}{3}

into

x^{3} + 2 x^{2} : \frac{32}{27}

Maximum (- \frac{4}{3}, \frac{32}{27})

Plug

x = 0

into

x^{3} + 2 x^{2} : 0

Minimum (0, 0)

Maximum (- \frac{4}{3}, \frac{32}{27}), Minimum (0, 0)

(5, 9), (6, - 1)

y = \frac{3 - x}{x + 1}

f (x) = (x + 4)^{2} - 9

f (x) = x (75 - 3 x)

\frac{x ^{2} - 3 x - 2}{x ^{2} - 4}

What is the extreme f(x)=x^3+2x^2 ?
The extreme f(x)=x^3+2x^2 is Maximum(-4/3 , 32/27),Minimum(0,0)