What is the extreme f(x)=(4x)/(x^2+1)+98.6 ?

The extreme f(x)=(4x)/(x^2+1)+98.6 is Minimum(-1,96.6),Maximum(1,100.6)

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extreme f(x)=(4x)/(x^2+1)+98.6

extreme points

f (x) = \frac{4 x}{x ^{2} + 1} + 98.6

Minimum (- 1, 96.6), Maximum (1, 100.6)

Solution steps

f^{'} (x) = - \frac{4 ( x + 1 ) ( x - 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} + 98.6 : 96.6

Minimum (- 1, 96.6)

Plug

x = 1

into

\frac{4 x}{x ^{2} + 1} + 98.6 : 100.6

Maximum (1, 100.6)

Minimum (- 1, 96.6), Maximum (1, 100.6)

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

f (x) = 10 x - 9

f (x) = 5 x - 2

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

y = \frac{x + 2}{x - 1}

What is the extreme f(x)=(4x)/(x^2+1)+98.6 ?
The extreme f(x)=(4x)/(x^2+1)+98.6 is Minimum(-1,96.6),Maximum(1,100.6)