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

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

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

extreme points

y = \frac{4 x}{x ^{2} + 1}

Minimum (- 1, - 2), Maximum (1, 2)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - 2)

Plug

x = 1

into

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

Maximum (1, 2)

Minimum (- 1, - 2), Maximum (1, 2)

f (x) = - 5 x^{6} ln (x), (0, 4)

f (x) = 8 sin (x) + 8 cos (x)

f (x) = 5 x + 5 sin (x), 0 \leq x \leq 2 π

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

f (x) = \frac{x ^{2}}{x ^{2} + 27}

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