What is the extreme f(x)=(6(x-7)^2)/(x+9) ?

The extreme f(x)=(6(x-7)^2)/(x+9) is Maximum(-25,-384),Minimum(7,0)

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extreme f(x)=(6(x-7)^2)/(x+9)

extreme points

f (x) = \frac{6 ( x - 7 ) ^{2}}{x + 9}

Maximum (- 25, - 384), Minimum (7, 0)

Solution steps

f^{'} (x) = \frac{6 ( x ^{2} + 18 x - 175 )}{( x + 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 25,

Decreasing

: - 25 < x < - 9,

Decreasing

: - 9 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 25

into

\frac{6 ( x - 7 ) ^{2}}{x + 9} : - 384

Maximum (- 25, - 384)

Plug

x = 7

into

\frac{6 ( x - 7 ) ^{2}}{x + 9} : 0

Minimum (7, 0)

Maximum (- 25, - 384), Minimum (7, 0)

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

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

f (x) = \frac{( x ^{9} )}{e ^{2 x}}

f (x, y) = x^{2} - y^{2} + 21

- \frac{5 e ^{4 x}}{x - 4}

What is the extreme f(x)=(6(x-7)^2)/(x+9) ?
The extreme f(x)=(6(x-7)^2)/(x+9) is Maximum(-25,-384),Minimum(7,0)