NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3)

* This is a calc problem

Question

NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3) 

* This is a calc problem

freckles · Answer

Have you considered finding the derivative?

usukidoll · Answer

we need the first derivative test... wait we need to find critical points first!

freckles · Answer

f'>0 tells us f is increasing 
f'<0 tells us f is decreasing 
Find the critical numbers and consider the domain.

usukidoll · Answer

what @freckles said is correct... use product rule

anonymous · Answer

I already did all that, and I got (sqrt x+3) + x/(2(sqrt x +3)). But I just dont know how to go about solving for the critical values

freckles · Answer

I don't think we need to find the x-intercepts to find where the function is increasing/decreasing.

anonymous · Answer

is it  just x= -3?

usukidoll · Answer

noooooo....................... freckles we need the critical points first or we can't go further T_T

usukidoll · Answer

sorry mary that comment was at freckles

freckles · Answer

$$f'(x)=\sqrt{x+3}+\frac{x}{2 \sqrt{x+3}} \ \text{ solve } f'(x)=0 \text{ \to find the critical numbers }$$

zale101 · Answer

To get the critical points, we first need to set the derivative to zero

anonymous · Answer

Yea I did that already guys!!!

zale101 · Answer

and solve for x. Those x's are the crtical points.

freckles · Answer

good then what did you for the x's?

anonymous · Answer

I just need help in solving for x

freckles · Answer

$$\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}$$
find common denominator

anonymous · Answer

No, we find the first derivative and then find the critical points

freckles · Answer

$$\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0$$

anonymous · Answer

hey freckles, how did you get +x in the numerator?

freckles · Answer

(sqrt x+3) + x/(2(sqrt x +3)
isn't this what you got for f'?

anonymous · Answer

OH, K i see it

freckles · Answer

$$f'(x)=\sqrt{x+3}+\frac{x}{ 2 \sqrt{x+3}}  \ f'(x) =\frac{ \sqrt{x+3} }{1} \cdot \frac{2 \sqrt{x+3}}{2 \sqrt{x+3}} +\frac{x}{2 \sqrt{x+3}} \ f'(x)=\frac{2(\sqrt{x+3})^2+x}{2 \sqrt{x+3}}$$

anonymous · Answer

So is it x=-2 for when f'(x) = 0

freckles · Answer

yes

anonymous · Answer

and then the denominator would give me when derivative is undefined?

anonymous · Answer

so the critical points are x= -2, and -3?

freckles · Answer

in by the way the domain of f is x>=-3 

so we have this so far:
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