What is the extreme f(x)=(x+1)/(x^2+8) ?

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

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extreme f(x)=(x+1)/(x^2+8)

extreme points

f (x) = \frac{x + 1}{x ^{2} + 8}

Minimum (- 4, - \frac{1}{8}), Maximum (2, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{- x ^{2} - 2 x + 8}{( x ^{2} + 8 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 4

into

\frac{x + 1}{x ^{2} + 8} : - \frac{1}{8}

Minimum (- 4, - \frac{1}{8})

Plug

x = 2

into

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

Maximum (2, \frac{1}{4})

Minimum (- 4, - \frac{1}{8}), Maximum (2, \frac{1}{4})

f (x) = \frac{8}{10 + x}

2 (x + 3)^{3} + 4

y = \frac{2}{5}

f (x) = - 2 (7)^{x}

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

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