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

The extreme f(x)=(x^2+12)(4-x^2) is Maximum(0,48)

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

extreme points

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

Maximum (0, 48)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

(x^{2} + 12) (4 - x^{2}) : 48

Maximum (0, 48)

4 x^{2} + 6 + \frac{6}{x ^{2} - 4 x + 4}

(x - y) (25 - x y)

f (x, y) = 2 x^{2} + 3 y^{2}

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

f (x) = - x^{4} - 8 x^{3} - 18 x^{2} + 7

What is the extreme f(x)=(x^2+12)(4-x^2) ?
The extreme f(x)=(x^2+12)(4-x^2) is Maximum(0,48)