What is the extreme f(x)=x-6sqrt(x+9) ?

The extreme f(x)=x-6sqrt(x+9) is Minimum(-9,-9),Minimum(0,-18)

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extreme f(x)=x-6sqrt(x+9)

extreme points

f (x) = x - 6 x + 9

Minimum (- 9, - 9), Minimum (0, - 18)

Solution steps

f^{'} (x) = 1 - \frac{3}{x + 9}

Find intervals:

Increasing

: - 9 \leq x < - \frac{80}{9},

Decreasing

: - \frac{80}{9} < x < \infty

Plug

x = - 9

into

x - 6 x + 9 : - 9

Minimum (- 9, - 9)

Plug

x = 0

into

x - 6 x + 9 : - 18

Minimum (0, - 18)

Minimum (- 9, - 9), Minimum (0, - 18)

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

x^{2} - 10 x - 9, 2 \leq x \leq 7

0.002 x^{2} + 4.4 x - 90

f (x) = 7 - 4 x^{2}

f (x, y) = \frac{( 1 x + 3 y )}{1 + x ^{2} + y ^{2}}

What is the extreme f(x)=x-6sqrt(x+9) ?
The extreme f(x)=x-6sqrt(x+9) is Minimum(-9,-9),Minimum(0,-18)