What is the extreme f(x)=(2(x+6)^2)/(x+8) ?

The extreme f(x)=(2(x+6)^2)/(x+8) is Maximum(-10,-16),Minimum(-6,0)

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extreme f(x)=(2(x+6)^2)/(x+8)

extreme points

f (x) = \frac{2 ( x + 6 ) ^{2}}{x + 8}

Maximum (- 10, - 16), Minimum (- 6, 0)

Solution steps

f^{'} (x) = \frac{2 ( x ^{2} + 16 x + 60 )}{( x + 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 10,

Decreasing

: - 10 < x < - 8,

Decreasing

: - 8 < x < - 6,

Increasing

: - 6 < x < \infty

Plug

x = - 10

into

\frac{2 ( x + 6 ) ^{2}}{x + 8} : - 16

Maximum (- 10, - 16)

Plug

x = - 6

into

\frac{2 ( x + 6 ) ^{2}}{x + 8} : 0

Minimum (- 6, 0)

Maximum (- 10, - 16), Minimum (- 6, 0)

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

\frac{5 x - 15}{2 x - 9}

f (x, y) = 12 x y - x^{2} - 3 y^{2}

f (x) = x^{3} - 12 x^{2} + 36 x - 3, 0 \leq x \leq 9

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

What is the extreme f(x)=(2(x+6)^2)/(x+8) ?
The extreme f(x)=(2(x+6)^2)/(x+8) is Maximum(-10,-16),Minimum(-6,0)