What is the extreme f(x)=xsqrt(x+9) ?

The extreme f(x)=xsqrt(x+9) is Maximum(-9,0),Minimum(-6,-6sqrt(3))

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extreme f(x)=xsqrt(x+9)

extreme points

f (x) = x x + 9

Maximum (- 9, 0), Minimum (- 6, - 63)

Solution steps

f^{'} (x) = \frac{3 x + 18}{2 x + 9}

Find intervals:

Decreasing

: - 9 < x < - 6,

Increasing

: - 6 < x < \infty

Plug

x = - 9

into

x x + 9 : 0

Maximum (- 9, 0)

Plug

x = - 6

into

x x + 9 : - 63

Minimum (- 6, - 63)

Maximum (- 9, 0), Minimum (- 6, - 63)

f (x) = 6 x^{\frac{2}{3}} - 4 x

f (x) = 3 x^{5} - 4 x^{3}

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

f (x, y) = 4 x^{2} + y^{2} - 8 x + 6 y + 8

f (x) = e^{x^{2} - 9 x - 1}, - 9 \leq x \leq 9

What is the extreme f(x)=xsqrt(x+9) ?
The extreme f(x)=xsqrt(x+9) is Maximum(-9,0),Minimum(-6,-6sqrt(3))