What is the extreme f(x)=(2x^2-10)/(x+3) ?

The extreme f(x)=(2x^2-10)/(x+3) is Maximum(-5,-20),Minimum(-1,-4)

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extreme f(x)=(2x^2-10)/(x+3)

extreme points

f (x) = \frac{2 x ^{2} - 10}{x + 3}

Maximum (- 5, - 20), Minimum (- 1, - 4)

Solution steps

f^{'} (x) = \frac{2 x ^{2} + 12 x + 10}{( x + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 5

into

\frac{2 x ^{2} - 10}{x + 3} : - 20

Maximum (- 5, - 20)

Plug

x = - 1

into

\frac{2 x ^{2} - 10}{x + 3} : - 4

Minimum (- 1, - 4)

Maximum (- 5, - 20), Minimum (- 1, - 4)

3600 + 118 x - 15 x^{2} + 0.5 x^{3}

3 x^{3} + \frac{144}{x}

\frac{2 + x - x ^{2}}{( x - 1 ) ^{2}}

y = 100 x (2 x + 3) (x - 5)

f (x) = 3 x^{2} + 2 y^{2} - 18 x + 4 y

What is the extreme f(x)=(2x^2-10)/(x+3) ?
The extreme f(x)=(2x^2-10)/(x+3) is Maximum(-5,-20),Minimum(-1,-4)