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

The extreme f(x)=x^3+12x^2 is Maximum(-8,256),Minimum(0,0)

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

extreme points

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

Maximum (- 8, 256), Minimum (0, 0)

Solution steps

f^{'} (x) = 3 x^{2} + 24 x

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 8

into

x^{3} + 12 x^{2} : 256

Maximum (- 8, 256)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 8, 256), Minimum (0, 0)

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

f (x) = \frac{x ^{2} - 16}{x + 5}

f (x) = x^{2} - 8 x + 13

f (x) = x^{\frac{2}{3}} - 4 x^{\frac{1}{3}}

f (x) = x^{2} - 6 x, - 1 \leq x \leq 6

What is the extreme f(x)=x^3+12x^2 ?
The extreme f(x)=x^3+12x^2 is Maximum(-8,256),Minimum(0,0)