What is the extreme f(x)=(x^2+12)(1-x^2) ?

The extreme f(x)=(x^2+12)(1-x^2) is Maximum(0,12)

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extreme f(x)=(x^2+12)(1-x^2)

extreme points

f (x) = (x^{2} + 12) (1 - x^{2})

Maximum (0, 12)

Solution steps

f^{'} (x) = - 2 x (2 x^{2} + 11)

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

(x^{2} + 12) (1 - x^{2}) : 12

Maximum (0, 12)

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

f (x) = \frac{- 3}{x ^{2} - 9}

f (x, y) = x y + \frac{9}{x} + \frac{3}{y}

6 x - x^{2}

f (x) = \frac{2 x ^{3}}{3} + 3 x^{2} - 20 x, (- 7, 5)

What is the extreme f(x)=(x^2+12)(1-x^2) ?
The extreme f(x)=(x^2+12)(1-x^2) is Maximum(0,12)