A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
find the intervals of increase and decrease for e^2xe^x, please help!
anonymous
 one year ago
find the intervals of increase and decrease for e^2xe^x, please help!

This Question is Closed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here we have to compute the first derivative of your function

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I got this: \[\Large f'\left( x \right) = {e^x}\left( {2{e^x}  1} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now the intervals of the real line in which f(x) is increasing, are given by the subsequent condition: \[\Large f'\left( x \right) > 0 \Rightarrow {e^x}\left( {2{e^x}  1} \right) > 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2which, since e^x is always positive, is equivalent to this one: \[\Large 2{e^x}  1 > 0\] please solve that inequality, what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please now solve it for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be ln(e)>1/2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, since if we take the logarithm of both sides, we get: \[\Large \begin{gathered} x\left( {\ln e} \right) > \ln \left( {1/2} \right) \hfill \\ x >  \ln 2 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please remember that: \[\Large \begin{gathered} \ln e = 1 \hfill \\ \ln \left( {1/2} \right) = \ln 1  \ln 2 = 0  \ln 2 =  \ln 2 \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay i see it now

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so your function is increasing in: [ln 2, +infinity) and decreasing, in: (infinity, ln2)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the domain of your function is all the real line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! and to find the local max and min i would have to find the second derivative and plug in ln2 for x right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, the local max and min comes from the first derivative

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now we have this: dw:1435213594909:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as we can see, point x=ln(2) is a local minimum for f(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I plug that into the first derivative so i can get the interval?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you can get the intervals in which your function is an increasing or a decreasing function, only solving the inequality: f '(x) >0 for example Now we have found x=ln(2) or, equivalently e^x= 1/2. At that point the value of your function is: \[\Large f\left( {  \ln 2} \right) = \frac{1}{4}  \frac{1}{2} =  \frac{1}{4}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so my max would be at (ln(2),1/4)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is a local minimum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait thats the min right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and theres no max for this function right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2that's right! there are no points of local maximum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for my concave up and concave down, would it be for concave down (ln(2),1/4) and for concave up (infinity,ln(2))U(ln(2),+infinity)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in order to establish the concavity type, we have to study the sign of the second derivative

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2after a simple computation, I got this: \[\large f''\left( x \right) = {e^x}\left( {4{e^x}  1} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, our function is concave up when f ''(x)>0, and it is concave down when f ''(x)<0 dw:1435214691996:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we have to solve this inequality: \[\Large f''\left( x \right) > 0 \Rightarrow {e^x}\left( {4{e^x}  1} \right) > 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next, since e^x is always positive, then that inequality is equivalent to this one: \[\Large 4{e^x}  1 > 0\] please solve it for x, what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we can write it as below: \[\Large x >  \ln 4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so your function, is concave down in this interval: (infinity, ln4), and concave up in this interval: [ln4, +infinity)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find the inflection point I would plug in ln4 into the second derivative and then that result back into the original equation right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait nvm i would solve for x in the second derivative?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, you have to plug in ln4 into the function, like this: \[\large f\left( {  \ln 4} \right) = {\left( {\frac{1}{4}} \right)^2}  \frac{1}{4} = \frac{1}{{16}}  \frac{1}{4} =  \frac{3}{{16}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so your inflection point is: \[\Large \left( {  \ln 4,  \frac{3}{{16}}} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay i understand now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much!! I really appreciated it!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.