What is the extreme f(x)=(x^2+7x+13)/(x+4) ?

The extreme f(x)=(x^2+7x+13)/(x+4) is Maximum(-5,-3),Minimum(-3,1)

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extreme f(x)=(x^2+7x+13)/(x+4)

extreme points

f (x) = \frac{x ^{2} + 7 x + 13}{x + 4}

Maximum (- 5, - 3), Minimum (- 3, 1)

Solution steps

f^{'} (x) = \frac{x ^{2} + 8 x + 15}{( x + 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 4,

Decreasing

: - 4 < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 5

into

\frac{x ^{2} + 7 x + 13}{x + 4} : - 3

Maximum (- 5, - 3)

Plug

x = - 3

into

\frac{x ^{2} + 7 x + 13}{x + 4} : 1

Minimum (- 3, 1)

Maximum (- 5, - 3), Minimum (- 3, 1)

\frac{( 5 x - 8 )}{x}

f (x) = sin (x), - 5 \leq x \leq 6

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

f (x) = 8 + (9 + 5 x)^{\frac{2}{5}}

f (x, y) = 7 x^{2} + 8 y^{2} + 4 x - 5 y + 8

What is the extreme f(x)=(x^2+7x+13)/(x+4) ?
The extreme f(x)=(x^2+7x+13)/(x+4) is Maximum(-5,-3),Minimum(-3,1)