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

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

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

extreme points

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

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

Solution steps

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

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

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 1

into

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

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

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

f (x) = (8 x)^{3} - (5 x)^{2} - 3 x

f (x) = 2

f (x, y) = 3 x^{3} + x y^{2} - 2 x y + 1

g (x) = x^{3} - 3 x^{2} + 3

f (x, y) = \frac{x ^{2} + y}{x y - 1}

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