What is the extreme f(x)=(3x^2)/(x+5) ?

The extreme f(x)=(3x^2)/(x+5) is Maximum(-10,-60),Minimum(0,0)

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extreme f(x)=(3x^2)/(x+5)

extreme points

f (x) = \frac{3 x ^{2}}{x + 5}

Maximum (- 10, - 60), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{3 ( x ^{2} + 10 x )}{( x + 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 10,

Decreasing

: - 10 < x < - 5,

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 10

into

\frac{3 x ^{2}}{x + 5} : - 60

Maximum (- 10, - 60)

Plug

x = 0

into

\frac{3 x ^{2}}{x + 5} : 0

Minimum (0, 0)

Maximum (- 10, - 60), Minimum (0, 0)

f (x) = 4 x^{2} + y^{2} - 12 x - 6 y - 2 x y + 3

f (x) = 2 y - 2 - x

f (x) = 9 - 8 x - x^{3}

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

f (x, y) = \frac{x ^{2} - 2 x y + y}{3 x - 2 y - 6}

What is the extreme f(x)=(3x^2)/(x+5) ?
The extreme f(x)=(3x^2)/(x+5) is Maximum(-10,-60),Minimum(0,0)