What is the extreme x/(x^2+16x+60) ?

The extreme x/(x^2+16x+60) is Minimum(-2sqrt(15),(4+sqrt(15))/4),Maximum(2sqrt(15),(4-sqrt(15))/4)

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extreme x/(x^2+16x+60)

extreme points

\frac{x}{x ^{2} + 16 x + 60}

Minimum (- 215, \frac{4 + 15}{4}), Maximum (215, \frac{4 - 15}{4})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 60}{( x ^{2} + 16 x + 60 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 10,

Decreasing

: - 10 < x < - 215,

Increasing

: - 215 < x < - 6,

Increasing

: - 6 < x < 215,

Decreasing

: 215 < x < \infty

Plug

x = - 215

into

\frac{x}{x ^{2} + 16 x + 60} : \frac{4 + 15}{4}

Minimum (- 215, \frac{4 + 15}{4})

Plug

x = 215

into

\frac{x}{x ^{2} + 16 x + 60} : \frac{4 - 15}{4}

Maximum (215, \frac{4 - 15}{4})

Minimum (- 215, \frac{4 + 15}{4}), Maximum (215, \frac{4 - 15}{4})

∣ sin (x) ∣

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 4

x^{3} + 2 x^{2} + x + 2

f (x) = x^{3} - 18 x^{2} + 81 x

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

What is the extreme x/(x^2+16x+60) ?
The extreme x/(x^2+16x+60) is Minimum(-2sqrt(15),(4+sqrt(15))/4),Maximum(2sqrt(15),(4-sqrt(15))/4)