What is the extreme f(x)=(3(x-4)^2)/(x+1) ?

The extreme f(x)=(3(x-4)^2)/(x+1) is Maximum(-6,-60),Minimum(4,0)

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extreme f(x)=(3(x-4)^2)/(x+1)

extreme points

f (x) = \frac{3 ( x - 4 ) ^{2}}{x + 1}

Maximum (- 6, - 60), Minimum (4, 0)

Solution steps

f^{'} (x) = \frac{3 ( x ^{2} + 2 x - 24 )}{( x + 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 1,

Decreasing

: - 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 6

into

\frac{3 ( x - 4 ) ^{2}}{x + 1} : - 60

Maximum (- 6, - 60)

Plug

x = 4

into

\frac{3 ( x - 4 ) ^{2}}{x + 1} : 0

Minimum (4, 0)

Maximum (- 6, - 60), Minimum (4, 0)

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

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

f (x) = 5 x + 10 a

f (x, y) = x^{2} + y^{2} - 20 x + 16 y - 9

4 - 6 x + x^{2}

What is the extreme f(x)=(3(x-4)^2)/(x+1) ?
The extreme f(x)=(3(x-4)^2)/(x+1) is Maximum(-6,-60),Minimum(4,0)