What is the extreme f(x)=2x^3-12x^2 ?

The extreme f(x)=2x^3-12x^2 is Maximum(0,0),Minimum(4,-64)

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extreme f(x)=2x^3-12x^2

extreme points

f (x) = 2 x^{3} - 12 x^{2}

Maximum (0, 0), Minimum (4, - 64)

Solution steps

f^{'} (x) = 6 x^{2} - 24 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

2 x^{3} - 12 x^{2} : 0

Maximum (0, 0)

Plug

x = 4

into

2 x^{3} - 12 x^{2} : - 64

Minimum (4, - 64)

Maximum (0, 0), Minimum (4, - 64)

f (x, y) = x^{3} - 3 x + y^{3} + \frac{5}{2} y^{2} - 2 y

f (x) = sin (x) cos (x), 0 \leq x \leq π

f (x) = 3 x^{3} + 18 x^{2} - 45 x + 26

f (x) = 4 x^{3} + 3 x^{2} - 12 x

2 x^{3} - 3 x^{2} - 36 x

What is the extreme f(x)=2x^3-12x^2 ?
The extreme f(x)=2x^3-12x^2 is Maximum(0,0),Minimum(4,-64)