# Find all asymptotes

1. Dec 9, 2016

### Starrrrr

1. Find all asymptotes of the graph of f(x).
f(x)= 1/1+e^3x

2. dy/dx= vdu/dx-udv/dx/v^2

3. Vertical asymptotes:
1+e^3x=0
3xIne=-1
X=-1/3
Horizontal asymptotes:
No horizontal asymptotes

Slant asymptotes:
None
Are the asymptotes I found correct?
Also I was wondering how can I get the first derivative of f(x) ?

Last edited by a moderator: Dec 9, 2016
2. Dec 9, 2016

### haruspex

How would you find horizontal asymptotes?
That step is wrong. Try it again. If you can't spot the error, try it in smaller steps.

3. Dec 9, 2016

### Staff: Mentor

Surely you don't mean $f(x) = \frac 1 1 + e^{3x}$, even though that's what you wrote.
As text, write this as f(x) = 1/(1 + e^(3x))

Also, if you intend to find the derivative, this is not a precalculus problem, so I'm moving it from the Precalc section to the Calculus & Beyond section.
Use parentheses!! This is NOT the quotient rule.
No, this is incorrect.
The most obvious way is to use the quotient rule. There's another way that might be simpler, that uses the chain rule.

4. Dec 9, 2016

### Starrrrr

I got it now , there are no vertical asymptotes. Horizontal asymptotes as x tends to infinity 1/(1+e^(3x)) = 1and negative infinity is 0

5. Dec 9, 2016

### Starrrrr

And also I have to use the chain rule

6. Dec 9, 2016

### Buffu

No, it is incorrect.
We know that $\huge{\lim_{x \to \infty}} 1 + e^{3x} = \infty$ and $\lim_{x \to -\infty} 1 + e^{3x} = 0$, So the positive horizontal asymptote should be 0 and negative horizontal asymptote is 1.