What is the extreme ((x+3))/(x^2-4) ?

The extreme ((x+3))/(x^2-4) is Minimum(-3-sqrt(5),-(3-sqrt(5))/8),Maximum(-3+sqrt(5),-(sqrt(5)+3)/8)

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extreme ((x+3))/(x^2-4)

extreme points

\frac{( x + 3 )}{x ^{2} - 4}

Minimum (- 3 - 5, - \frac{3 - 5}{8}), Maximum (- 3 + 5, - \frac{5 + 3}{8})

Solution steps

f^{'} (x) = \frac{- x ^{2} - 6 x - 4}{( x ^{2} - 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3 - 5,

Increasing

: - 3 - 5 < x < - 2,

Increasing

: - 2 < x < - 3 + 5,

Decreasing

: - 3 + 5 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 3 - 5

into

\frac{x + 3}{x ^{2} - 4} : - \frac{3 - 5}{8}

Minimum (- 3 - 5, - \frac{3 - 5}{8})

Plug

x = - 3 + 5

into

\frac{x + 3}{x ^{2} - 4} : - \frac{5 + 3}{8}

Maximum (- 3 + 5, - \frac{5 + 3}{8})

Minimum (- 3 - 5, - \frac{3 - 5}{8}), Maximum (- 3 + 5, - \frac{5 + 3}{8})

f (x) = - 2 x - 2

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

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

f (x) = \frac{0.1 x}{x ^{2} + 16}

f (x) = - 9 cos (- \frac{π}{2} x - 6) + 8

What is the extreme ((x+3))/(x^2-4) ?
The extreme ((x+3))/(x^2-4) is Minimum(-3-sqrt(5),-(3-sqrt(5))/8),Maximum(-3+sqrt(5),-(sqrt(5)+3)/8)