Calculus Help

[B.def]

Ok, I give up. I need to find where the function is increasing and decreasing.
The function is... f(x)=e^(9x)+e^(-x) .... f'(x)=9e^(9x)-e^(-x)
My problem is solving for where the function equals zero. Any help would be greatly appreciated.

Also here is another problem I'm having the same issue.
f(x)=x^(5)*(lnx) f'(x)=x^4+lnx(5x^4)

 

[Fritzo]

 

[coxmaster]

Find where f'(x)=0

For the first one its pretty simple. Take a natural log of the right side, then divide through to get x.

0=9e^(9x)-e^(-x)
0=ln(9)+ln(e^(9x))-ln(e^(-x))
0=ln(9)+9x+x
0=ln(9)+10x

etc

 

[corwin]

Try this site...if you can type it in correctly it will give you the answer you seek

http://www.mathway.com/problem.aspx?p=calculus

 

[B.def]

Find where f'(x)=0

For the first one its pretty simple. Take a natural log of the right side, then divide through to get x.

0=9e^(9x)-e^(-x)
0=ln(9)+ln(e^(9x))-ln(e^(-x))
0=ln(9)+9x+x
0=ln(9)+10x

etc

That was my first thought, but then you are taking the natural log of 0, which is not possible.

Try this site...if you can type it in correctly it will give you the answer you seek

http://www.mathway.com/problem.aspx?p=calculus

I tried that, but only gives a decimal value, just like WolframAlpha... which I believe is useless in this situation.

Come on Calc Professors, where you at?! lol

 

[coxmaster]

That was my first thought, but then you are taking the natural log of 0, which is not possible.

Ah right.. Totally overlooked that.

Still shouldn't be difficult to manipulate it so itll work.

Try this
9e^(9x)-e^(-x)=0
Becomes
9e^(9x)=e^(-x)


You could also factor it out, that way you can set each of the terms equal to zero.

 

[B.def]

Ah right.. Totally overlooked that.

Still shouldn't be difficult to manipulate it so itll work.

Try this
9e^(9x)-e^(-x)=0
Becomes
9e^(9x)=e^(-x)


You could also factor it out, that way you can set each of the terms equal to zero.

9e^(9x)-e^(-x)=0
Becomes
9e^(9x)=e^(-x)
ln(9e^(9x))=ln(e^(-x)
9*9x=-x
81x=-x
81x+x=0
x(81+1)=0
x=0 ???

I must be doing something wrong still.

 

[coxmaster]

9e^(9x)-e^(-x)=0
Becomes
9e^(9x)=e^(-x)
ln(9e^(9x))=ln(e^(-x)
9*9x=-x
81x=-x
81x+x=0
x(81+1)=0
x=0 ???

I must be doing something wrong still.

Shit, you're right. That isn't going to work either. It looks like factoring it out is the only way to go. (Wolfram confirms this)

 

[B.def]

I hate calculus..... lol
but thanks for you help man

 

[SheHateMe]

WolframAlpha.com You're welcome

I would help you but my eyes have been staring at screens to long at work and it hurts to look at this equation the way you typed it.

 

[darkxshade]

u + me = us

 

[ringtail]

``

 

[SaurusX]

f'(x)=9e^(9x)-e^(-x)
l(x)=ln9+ln(e^(9x))-ln(e^(-x))
=ln9+9x+x, now you can set to 0
0=ln9+10x
-10x=ln9
x=-.219722