How Derivatives affect graphs

[h20skier]

Hello All, This problem comes from #18 of section 4.3 out of stewarts 6th edition.

The Question:

Consider the following function:
f(x)= e^(-x) * sqrt(x)

What I need.

a.) Find the intervals on which f is increasing and decreasing.
b.) find any relative extrema.
c.) find the intervals of concavity.
d.) find all inflection points.

For part a I have taken the first derivative and got. 1/2x^^-(1/2) *e^-x - e^-x * sqrt (x)

after simplifying i got e^-x= 0
and 1/2 x^-1/2 - sqrt x = 0

I know i need to find these for my critical points but I am having difficulty proceeding.

for extrema i know i plug in values between my critical points to see the critcal values.

after that I know I take the 2nd derivative to find critical points for my concavity.

thanks for your help!!

 

[pka]

Consider the following function:
f(x)= e^(-x) * sqrt(x)
a.) Find the intervals on which f is increasing and decreasing.
b.) find any relative extrema.
c.) find the intervals of concavity.
d.) find all inflection points.

\(\displaystyle y' = e^{ - x} \left( {\frac{1}{{2\sqrt x }} - \sqrt x } \right)\)

Now take note that \(\displaystyle \left( {\forall x} \right)\left[ {e^{ - x} \ne 0} \right]\).

Therefore, you must solve \(\displaystyle \left( {\frac{1}{{2\sqrt x }} - \sqrt x } \right) = 0\).

 

[mmm4444bot]

I [took] the first derivative and [got] 1/2x^^(-1/2)*e^(-x) - e^(-x)*sqrt(x)

after simplifying i got e^(-x)= 0

and 1/2 x^(-1/2) - sqrt(x) = 0

I am having difficulty proceeding.

At this point, you're trying to solve for x, yes?

You should know by inspection that the first equation has no solutions (think about the graph of y=e^x).

On the second equation, add sqrt(x) to both sides (to separate the radicals), and then square both sides. If you need more help, please show your work. :cool: