$$f(x) =\dfrac{e^x}{1+e^x}$$ I know we can find points of inflection using the second derivative test. The second derivative for the function above is $$f''(x) = \dfrac{e^x(1-e^x)}{(e^x+1)^3}$$ I have found one critical point for the second derivative which is $0$. I then determined that the function is concave up from $(-\infty,0)$ and concave down from $(0,\infty)$. I am now asked to find the points of inflection. How would I determine the exact points from where the function switches from concave up to concave down?

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    $\begingroup$ You've already shown that it switches from concave up to concave down at $0$. $\endgroup$ – Alex Becker May 24 '13 at 0:40
  • $\begingroup$ When the second derivative is zero AND switches sign, that's where you have an inflection point $\endgroup$ – imranfat May 24 '13 at 0:41
  • $\begingroup$ Oh... I must have misread the solution in the solution book. I reread it and found the inflection point is indeed at $(0,1/2)$. $\endgroup$ – Kot May 24 '13 at 0:44

You've found the inflection point by identifying the value of $x$ at which the graph shifts from concave up, to concave down. Plus, it matches the solution to the $f''(x) = 0$.

That gives you an inflection point at $\left(0, \frac 12\right)$,

  • $\begingroup$ when you say plus it matches the solution to the $f''(x)=0$ do you mean it matches the critical point? $\endgroup$ – Kot May 24 '13 at 0:55
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    $\begingroup$ Yes, exactly: the critical point of the second derivative test is the solution(s) to $f''(x) = 0$: where the second derivative equals $0$, if anywhere. In this case that's when $x = 0$. $\endgroup$ – Namaste May 24 '13 at 0:58

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