What is the extreme f(x)=xsqrt(3+x) ?

The extreme f(x)=xsqrt(3+x) is Maximum(-3,0),Minimum(-2,-2)

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extreme f(x)=xsqrt(3+x)

extreme points

f (x) = x 3 + x

Maximum (- 3, 0), Minimum (- 2, - 2)

Solution steps

f^{'} (x) = \frac{3 x + 6}{2 3 + x}

Find intervals:

Decreasing

: - 3 < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 3

into

x 3 + x : 0

Maximum (- 3, 0)

Plug

x = - 2

into

x 3 + x : - 2

Minimum (- 2, - 2)

Maximum (- 3, 0), Minimum (- 2, - 2)

F (x, y) = x^{2} + x^{2} y + y^{2} + 1

f (x) = - \frac{5 x}{4 x ^{2} + 7}

f (x) = e^{- 1.5 x^{2}}

f (x) = x^{3} - 9 x^{2} + 24 x - 2

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 7 x + 11 y

What is the extreme f(x)=xsqrt(3+x) ?
The extreme f(x)=xsqrt(3+x) is Maximum(-3,0),Minimum(-2,-2)