What is the extreme f(x)=(7(x-1)^2)/(x+9) ?

The extreme f(x)=(7(x-1)^2)/(x+9) is Maximum(-19,-280),Minimum(1,0)

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extreme f(x)=(7(x-1)^2)/(x+9)

extreme points

f (x) = \frac{7 ( x - 1 ) ^{2}}{x + 9}

Maximum (- 19, - 280), Minimum (1, 0)

Solution steps

f^{'} (x) = \frac{7 ( x ^{2} + 18 x - 19 )}{( x + 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 19,

Decreasing

: - 19 < x < - 9,

Decreasing

: - 9 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 19

into

\frac{7 ( x - 1 ) ^{2}}{x + 9} : - 280

Maximum (- 19, - 280)

Plug

x = 1

into

\frac{7 ( x - 1 ) ^{2}}{x + 9} : 0

Minimum (1, 0)

Maximum (- 19, - 280), Minimum (1, 0)

2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}}

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

\frac{d}{d x} ((x + y) 2 + 2 y^{2})

f (x) = \frac{2 ( x + 7 ) ^{2}}{x + 10}

\frac{2 x ^{2} - 5 x + 5}{x - 2}

What is the extreme f(x)=(7(x-1)^2)/(x+9) ?
The extreme f(x)=(7(x-1)^2)/(x+9) is Maximum(-19,-280),Minimum(1,0)