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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x (x^{2} - 12) : 16

Maximum (- 2, 16)

Plug

x = 2

into

x (x^{2} - 12) : - 16

Minimum (2, - 16)

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

f (x, y) = (x + y^{2}) \cdot e^{2 x}

f (x, y) = 8 e^{(\frac{( - x ^{2} - y ^{2} )}{4})}

f (x, y) = ln (16 - x^{2} - 16 y^{2})

f (x) = \frac{1}{3} x^{3} + \frac{13}{2} x^{2} + 42 x + 3

(x^{2} + 2 x) e^{- x}

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