What is the extreme f(x)=4x^3-12x^2 ?

The extreme f(x)=4x^3-12x^2 is Maximum(0,0),Minimum(2,-16)

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extreme f(x)=4x^3-12x^2

extreme points

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

Maximum (0, 0), Minimum (2, - 16)

Solution steps

f^{'} (x) = 12 x^{2} - 24 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 2

into

4 x^{3} - 12 x^{2} : - 16

Minimum (2, - 16)

Maximum (0, 0), Minimum (2, - 16)

f (x, y) = \frac{1}{x + y}

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

f (x, y) = x - 9 y + 17

\frac{x ^{2} - 2 x + 2}{x - 1}

f (x, y) = y^{3} + x^{2} - 6 x y + 3 x + 6 y - 7

What is the extreme f(x)=4x^3-12x^2 ?
The extreme f(x)=4x^3-12x^2 is Maximum(0,0),Minimum(2,-16)