critical point of (x-1)e^x

[dubb]

Hi, I'm trying to locate the critical points of f(x)=(x-1)e^x Using the first derivative test to find local extrema. I understand how to do most of the process but I get stuck on the derivative when it includes e^x. Could you please help?

Here's what I have but I dont thin
I been able to figure out that the derivative is:
f'= e^x+e^x(x-1)

I'm getting stuck when finding critical points by making making f'=0,
e^x+e^x(x-1)=0
does this equal x=0, 0 is the only critical point?

Thanks.

 

[Deleted member 4993]

Hi, I'm trying to locate the critical points of f(x)=(x-1)e^x Using the first derivative test to find local extrema. I understand how to do most of the process but I get stuck on the derivative when it includes e^x. Could you please help?

Here's what I have but I dont thin
I been able to figure out that the derivative is:
f'= e^x+e^x(x-1)

I'm getting stuck when finding critical points by making making f'=0,
e^x+e^x(x-1)=0
does this equal x=0, 0 is the only critical point?

Thanks.

Before doing this problem, you should be very clear about the definition/characteristics of critical points of a function.

What is the definition/characteristics of critical points of a function?

 

[HallsofIvy]

Did you notice that \(\displaystyle e^x+ e^x(x- 1)= e^x+ xe^x- e^x= xe^x\)? Or is your function something else?

 

[dubb]

Before doing this problem, you should be very clear about the definition/characteristics of critical points of a function.

What is the definition/characteristics of critical points of a function?

Off the top of my head its where there can be extremas, or local extremas, i've also noticed its where the slope of tanget line is 0.

 

[dubb]

Did you notice that \(\displaystyle e^x+ e^x(x- 1)= e^x+ xe^x- e^x= xe^x\)? Or is your function something else?

Hi Hallso, Yes i noticed that, to be honest, for some reason when ever I see "e" in problem i get freaked out, i've actually been trying to focus on those problems to overcome that. I worked on the problem a little more:


e^x+ xe^x- e^x= xe^x (i'm not sure about what you wrote "tex") if i do (x)(e^x)=0 then i get 1 and 0 for critical points. does that look right?

 

[pka]

e^x+ xe^x- e^x= xe^x (i'm not sure about what you wrote "tex") if i do (x)(e^x)=0 then i get 1 and 0 for critical points. does that look right?

That is incorrect \(\displaystyle xe^x=0\) means that \(\displaystyle x=0\).

Now \(\displaystyle f''(0)=~?\)

 

[dubb]

That is incorrect \(\displaystyle xe^x=0\) means that \(\displaystyle x=0\).

Now \(\displaystyle f''(0)=~?\)

f'' would be e^x+xe^x --> e^x(1+x)
I used the product rule.
look right?

 

[lookagain]

Before doing this problem, you should be very clear about the definition/characteristics of critical points of a function.

What is the definition/characteristics of critical points of a function?

Off the top of my head its where there can be extremas, or local extremas, i've also noticed its where the slope of tanget line is 0.

dubb,
"critical numbers are values of x at which

or

does not exist."


Source:
http://sites.csn.edu/istewart/mathweb/Math181/critical_numb/critical_numb.htm


- -- - - - - - - - - --- - - -- - - - -- - - - - ---- - - - - - - - --- - - - - - - - - - - -- - - - - - - - - - - - - - ---- -

Using the first derivative test to find local extrema.

xe^x = 0

x = 0 \(\displaystyle \ \ \ \ \ \) e^x = 0 has no solution for real x.


x is the only critical number here. To find the critical point,, substitute x into the original function to get
the y-value.


If you stay with the first derivative test, pick a convenient number less x = 0 and a convenient number
greater than x = 0 and substitute each one into the first derivative. As x increases about 0, if the sign
of the derivative changes from negative to positive, then it corresponds to a relative minimum.
As x increases about 0, if the sign of the derivative changes from positive to negative, then it corresponds
to a relative maximum.


Suggestion: Test x = -1, and then test x = 1 in the first derivative.