$$f(x) =\frac{1}{1+e^{-2x}}$$

Find all asymptotes of the graph of f(x).

Determine the interval on which f(x) is decreasing.

Determine the interval on which f(x) is increasing.

Determine all the points of inflection of f(x)

The asymptotes are vertical none , horizontal 0 and 1 then to find the interval decreasing , increasing you will have to get first derivative but I do use the chain rule?

  • $\begingroup$ You only need derivatives for the inflection point. To determine monotonicity it's enough to note that $e^{-2x}$ is positive and decreasing, so $1+e^{-2x}$ is positive and decreasing, so $1 / (1+e^{-2x)}$ is positive and increasing. $\endgroup$ – dxiv Dec 11 '16 at 3:29

Let's simplify $f(x)$ a bit.

$$f(x) =\frac{1}{1+e^{-2x}} = \frac{1}{\left(1+\frac{1}{e^{2x}}\right)}$$ $$ = \frac{1}{\frac{\left(e^{2x}+1\right)}{e^{2x}}}$$ $$ = \frac{e^{2x}}{e^{2x}+1}$$

Now this is something that you can differentiate using chain rule/quotient rule.

$$f'(x) = \frac{e^{2x}\left(2\right)\left(e^{2x}+1\right)-e^{2x}\left(e^{2x}\right)\left(2\right)}{\left(e^{2x}+1\right)^2} $$ $$ f'(x) = \frac{2e^{2x}\left(e^{2x}+1-e^{2x}\right)}{\left(e^{2x}+1\right)^2}$$ $$ f'(x) = \frac{2e^{2x}}{\left(e^{2x}+1\right)^2}$$

What we notice here is that $f'(x)$ is never zero, and thus it has $no$ critical points. So, if you plug in any random value, you will see that the second deriative is $always$ positive, and thus always $increasing$.

Then you find $f''(x)$ for inflection points.

  • $\begingroup$ So when I find the second derivative I factorise it to find the inflection points? Thank you $\endgroup$ – user373675 Dec 11 '16 at 3:47
  • $\begingroup$ When you find the second derivative, you set it equal to zero, and whatever points you get, those might be inflection. Then you can check to see if the concavity changes, and if it does, then it is a inflection point. $\endgroup$ – K Split X Dec 11 '16 at 3:48
  • $\begingroup$ ok thank you , I really appreciate the help $\endgroup$ – user373675 Dec 11 '16 at 3:50
  • $\begingroup$ Your welcome! :) $\endgroup$ – K Split X Dec 11 '16 at 3:50
  • $\begingroup$ I didnt even know that , thank you so much for the clear explanation :) $\endgroup$ – user373675 Dec 11 '16 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy