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

The extreme f(x)=2((x^3)/3+x^2) is Maximum(-2, 8/3),Minimum(0,0)

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

extreme points

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

Maximum (- 2, \frac{8}{3}), Minimum (0, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

2 (\frac{x ^{3}}{3} + x^{2}) : \frac{8}{3}

Maximum (- 2, \frac{8}{3})

Plug

x = 0

into

2 (\frac{x ^{3}}{3} + x^{2}) : 0

Minimum (0, 0)

Maximum (- 2, \frac{8}{3}), Minimum (0, 0)

f (x) = \frac{x}{x ^{2} - 49}

7 x (1 + 12 x^{2})^{- 2}

f (x) = 4 sin (x) + 3 cos (x), 0 \leq x \leq 2 π

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

5 x^{2} (x + 1)^{\frac{1}{2}}

What is the extreme f(x)=2((x^3)/3+x^2) ?
The extreme f(x)=2((x^3)/3+x^2) is Maximum(-2, 8/3),Minimum(0,0)