Question

find the critical numbers and the open intervals on which the function is increasing or decreasing...

Answer 1

\[f(x)=x \sqrt{x+1}\]

Answer 2

google it

Answer 3

can you differentiate f(x) ?

Answer 4

you'll need the derivative of f(x) and to find the critical numbers, solve f'(x)=0.

Answer 5

do you have the derivative?

Answer 6

yes i got\[\frac{ 1 }{ 2 }x (x+1)^{-\frac{ 1 }{ 2 }}\]

is that correct? @byteme

Answer 7

no... that's not right... you'll need to use the product rule/chain rule....

Answer 8

\(\large f(x)=x\sqrt{x+1}=x(x+1)^{\frac{1}{2}} \)
\(\large f'(x)=[x]' \cdot (x+1)^{\frac{1}{2}}+x\cdot [(x+1)^{\frac{1}{2}}]' \)
= \(\large (x+1)^{\frac{1}{2}} + x \cdot \frac{1}{2}(x+1)^{-\frac{1}{2}} \cdot [x]' \)
= \(\large (x+1)^{\frac{1}{2}} + x \cdot \frac{1}{2}(x+1)^{-\frac{1}{2}} \)
= \(\large (x+1)^{-\frac{1}{2}}((x+1)^1 + \frac{x}{2}) \)
= \(\large \frac{\frac{3x}{2}+1}{(x+1)^{\frac{1}{2}}} \)

Answer 9

so \(\large f'(x)=0 \Rightarrow x=-\frac{3}{2} \)