What is the extreme f(x)=(2x^2)/(x^4+1) ?

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

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extreme f(x)=(2x^2)/(x^4+1)

extreme points

f (x) = \frac{2 x ^{2}}{x ^{4} + 1}

Maximum (- 1, 1), Minimum (0, 0), Maximum (1, 1)

Solution steps

f^{'} (x) = \frac{2 ( - 2 x ^{5} + 2 x )}{( x ^{4} + 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 1)

Plug

x = 0

into

\frac{2 x ^{2}}{x ^{4} + 1} : 0

Minimum (0, 0)

Plug

x = 1

into

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

Maximum (1, 1)

Maximum (- 1, 1), Minimum (0, 0), Maximum (1, 1)

\frac{\frac{7}{10}}{\frac{- 5}{10}}

2 x^{2} - 3 x + 1

8 x - 7 y = 15

f (x) = \frac{1}{x}

f (x) = \frac{2}{3} - 1

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