What is the extreme f(x)=(x^2)/(x^4+16) ?

The extreme f(x)=(x^2)/(x^4+16) is Maximum(-2, 1/8),Minimum(0,0),Maximum(2, 1/8)

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extreme f(x)=(x^2)/(x^4+16)

extreme points

f (x) = \frac{x ^{2}}{x ^{4} + 16}

Maximum (- 2, \frac{1}{8}), Minimum (0, 0), Maximum (2, \frac{1}{8})

Solution steps

f^{'} (x) = \frac{- 2 x ^{5} + 32 x}{( x ^{4} + 16 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{2}}{x ^{4} + 16} : \frac{1}{8}

Maximum (- 2, \frac{1}{8})

Plug

x = 0

into

\frac{x ^{2}}{x ^{4} + 16} : 0

Minimum (0, 0)

Plug

x = 2

into

\frac{x ^{2}}{x ^{4} + 16} : \frac{1}{8}

Maximum (2, \frac{1}{8})

Maximum (- 2, \frac{1}{8}), Minimum (0, 0), Maximum (2, \frac{1}{8})

f (x) = - 2 x^{3} + 6 x^{2} + 12

(x - 2)^{3} \cdot (x + 2)

f (x) = \frac{x + 2}{x ^{2} - 7 x - 18}

5 x y - 7 x^{2} + 3 x - 6 y + 2

f (x) = 3 x^{2} - 6 x + 2

What is the extreme f(x)=(x^2)/(x^4+16) ?
The extreme f(x)=(x^2)/(x^4+16) is Maximum(-2, 1/8),Minimum(0,0),Maximum(2, 1/8)