Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Goten77
Group Title
y=xe^(x/8)
y1=x(e^(x/8))(1/8) + e^(x/8)(1)
critical points
ima type this below along w/ the question
 2 years ago
 2 years ago
Goten77 Group Title
y=xe^(x/8) y1=x(e^(x/8))(1/8) + e^(x/8)(1) critical points ima type this below along w/ the question
 2 years ago
 2 years ago

This Question is Closed

Goten77 Group TitleBest ResponseYou've already chosen the best response.0
\[y=xe ^{x/8}\] then derivative is this correct? \[y ^{1}=xe ^{x/8}(1/8) + e ^{x/8}1\]
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
simplifying your 1st derivative you will have \[\frac{dy}{dx} = e^{\frac{x}{8}}(\frac{1}{8} x + 1)\] so you need to solve \[\frac{1}{8} x + 1 = 0\]
 2 years ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.0
ok thats what i got and then i get like 1/8x = 1 then x=1/(1/8) = +8
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
thats correct.... that is the only value of x that makes the 1st derivative = 0...
 2 years ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.0
oh nvm i c y e^o=1 eh i aint thinking
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
if you have x = 0 then \[dy/dx = e^{\frac{0}{8}}(\frac{1}{8}\times 0 + 1) = 1\] because \[e^{\frac{0}{8}} = 1\] any number to a power of zero is 1 so only 1 critical point...
 2 years ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.0
man i made so many simple mistakes on this last quiz... today i just aint thinking... oh well atleast i got this concept right XD
 2 years ago
See more questions >>>
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.