What is the extreme (2x+1)/(x^2+12x+48) ?

The extreme (2x+1)/(x^2+12x+48) is Minimum(-7,-1),Maximum(6, 1/12)

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extreme (2x+1)/(x^2+12x+48)

extreme points

\frac{2 x + 1}{x ^{2} + 12 x + 48}

Minimum (- 7, - 1), Maximum (6, \frac{1}{12})

Solution steps

f^{'} (x) = \frac{- 2 x ^{2} - 2 x + 84}{( x ^{2} + 12 x + 48 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = - 7

into

\frac{2 x + 1}{x ^{2} + 12 x + 48} : - 1

Minimum (- 7, - 1)

Plug

x = 6

into

\frac{2 x + 1}{x ^{2} + 12 x + 48} : \frac{1}{12}

Maximum (6, \frac{1}{12})

Minimum (- 7, - 1), Maximum (6, \frac{1}{12})

f (x) = x^{3} - 2 x^{2} + 3, (2, 7)

2 cos (x) - 2

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

f (y) = 6 - 3 x - 2 y

y^{2} - 120 y

What is the extreme (2x+1)/(x^2+12x+48) ?
The extreme (2x+1)/(x^2+12x+48) is Minimum(-7,-1),Maximum(6, 1/12)