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

The extreme f(x)=x^3+x^2-12 is Maximum(-2/3 , 4/27-12),Minimum(0,-12)

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

extreme points

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

Maximum (- \frac{2}{3}, \frac{4}{27} - 12), Minimum (0, - 12)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} + x^{2} - 12 : \frac{4}{27} - 12

Maximum (- \frac{2}{3}, \frac{4}{27} - 12)

Plug

x = 0

into

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

Minimum (0, - 12)

Maximum (- \frac{2}{3}, \frac{4}{27} - 12), Minimum (0, - 12)

f (x) = x^{4} - 72 x^{2} + 1296, - 7 \leq x \leq 7

f (x) = e^{- a t}

\frac{6 - x}{x ^{3}}

f (x, y) = 3 x y - 4 x - 4 y + 3

h (x) = 5 - x^{2} (- 3.1)

What is the extreme f(x)=x^3+x^2-12 ?
The extreme f(x)=x^3+x^2-12 is Maximum(-2/3 , 4/27-12),Minimum(0,-12)