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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 2

into

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

Maximum (2, 16)

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

f (x) = \frac{8 x}{x ^{2} + 16}

\frac{( x ^{2} - 1 )}{( x - 3 ) ^{2}}

s (t) = - 4.9 t^{2} + 29 t + 2

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

f (x) = 3 x - \frac{1}{2} x^{2} - \frac{1}{2} ln (x)

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