What is the extreme f(x)=2x^3+4x^2 ?

The extreme f(x)=2x^3+4x^2 is Maximum(-4/3 , 64/27),Minimum(0,0)

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extreme f(x)=2x^3+4x^2

extreme points

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

Maximum (- \frac{4}{3}, \frac{64}{27}), Minimum (0, 0)

Solution steps

f^{'} (x) = 6 x^{2} + 8 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

2 x^{3} + 4 x^{2} : \frac{64}{27}

Maximum (- \frac{4}{3}, \frac{64}{27})

Plug

x = 0

into

2 x^{3} + 4 x^{2} : 0

Minimum (0, 0)

Maximum (- \frac{4}{3}, \frac{64}{27}), Minimum (0, 0)

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

- lo g_{5} (x)

8 x^{2} + 14 x y + 3 y^{2} + 10 x - 4

f (x) = 6 x^{2} + 6 x - 12

f (x, y) = x^{3} - 2 x y + y^{2} + 4

What is the extreme f(x)=2x^3+4x^2 ?
The extreme f(x)=2x^3+4x^2 is Maximum(-4/3 , 64/27),Minimum(0,0)