Critical points and inflection points Is it always so that a point of inflection is also a critical point?
What about this function?:
$f(x)=2e^x/(1+e^x)$
 A: It is not always so that a point of inflection is also a critical point.
A critical point of a function is a point where the first derivative is undefined or zero. This is important because a minimum or maximum of a function defined on an interval must occur at an endpoint of the interval or at a critical point.
An inflection point of a function is a point where the concavity of the function changes. More simply, the second derivative is positive on one side of the point and negative on the other side. The second derivative at the inflection point is either undefined or zero. This is important since it tells you where the function is "changing direction": from curving up to curving down or the other way round.
You see that the critical points depend on the first derivative, while inflection points depend on the second derivative. There is very little relation between the two.

In your particular case, the first derivative is never zero, so there are no critical points. The second derivative is zero at $x=0$, is positive to the left of zero and negative to the right of zero. That means there is an inflection point at $x=0$.
Ask if you need more details.
A: Definition. An inflection point is a point on a curve $f(x)$ at which the curve changes from being concave (concave downward) to convex (concave upward), or vice versa. On the other hand, a critical point is a point for which $f'(x)=0$.
Concave downwards


*

*Slope is decreasing: $f'(x)$ is decreasing.

*$f''(x)<0$.
Concave upwards


*

*Slope is increasing: $f'(x)$ is increasing.

*$f''(x)>0$.
From your question, $f'(x)=2e^x(1+e^x)^{-1}-2e^{2x}(1+e^x)^{-2}$. Can you now find points of inflection and cricital points from here?
A: $$f'=\frac{2e^x(1+e^x)-e^x(2e^x))}{(1+e^x)^2}=\\ \frac{2e^x}{(1+e^x)^2} =0\\no-root \\$$so this function has no critical point ,note: domain is $\mathbb{R}$ $$f''=\frac {2e^x(1+e^x)^2-2(1+e^x)e^x(2e^x)}{(1+e^x)^4}=\\ \frac{2e^x(1-e^x)}{(1+e^x)^3}=0 \rightarrow e^x(1-e^x)=0 \\e^x=0 \space no-root\\1-e^x=0 \\e^x=1\\x=0 $$so $x=0$ is inflection point
  this inflection point is not critical ,but some of inflection  points are critical like these $$f(x)=(x-a)^{2n+1}g(x) ,\space n \in \mathbb{N} \\x=a \space \space is \space inflection \space point ,also \space critical \space because f'(a)=0$$ or this form $$ f(x)=\sqrt[2n+1]{(x-a)^{2k+1}}g(x) , n>k \in \mathbb{N}\\ \space x=a \space vertical \space \space inflection \space  point \space also \space critical \space because \space f'(a) \space does \space not \space exist $$
