Question

f is the real valued function defined by f(x) = 1 / (1 + e^x).
(i) Draw the variation table of f and sketch its representative curve.
(ii) Show that the point I(0, 1/2) is the center of symmetry of the curve of f.

Answer 1

(i)

(ii)

By the definition, let $f(x)$  be a real-valued function of a real variable. Then $f(x)$  is odd if the following equation holds for all $x$  such that $x$  and $(-x)$  are in the domain of $f(x)$ :

$$f(x)=-f(-x)$$

Geometrically, the graph of an odd function has rotational symmetry with respect to the origin.

(More information: https://en.wikipedia.org/wiki/Even_and_odd_functions)

To prove that the point $p.I(a,b)$ is the center of symmetry of the function $f(x)$ , it suffices to prove that the function $g(x)=f(x+a)-b$ is odd.

In our case, we want to prove that point $p.I\left(0,1/2\right)$ is the center of symmetry, so we need to consider the function

$$f(x)=\frac{1}{1+e^x}\to g(x)=f(x+0)-\frac{1}{2}=\frac{1}{1+e^x}^{/2}-\frac{1}{2}^{/ \left(1+e^x\right)}=\\[0.3cm] =\frac{2-1-e^x}{1+e^x}=\frac{1-e^x}{1+e^x}\to\boxed{g(x)=\frac{1-e^x}{1+e^x}}$$

It remains to prove that the new function $g(x)$ is odd.

$$g(-x)=\frac{1-e^{-x}}{1+e^{-x}}=\frac{e^{-x}\left(e^x-1\right)}{e^{-x}\left(e^x+1\right)}= \frac{-\left(1-e^x\right)}{1+e^x}=-g(x)\\[0.3cm] \boxed{g(-x)=-g(x)\to g(x)-\text{odd function}}$$

Conclusion,

Since $g(x)$ turned out to be an odd function, it means that its center of symmetry is the origin of coordinates $O(0,0)$ . This means that the function $f(x)$ has a center of symmetry, point $I(0,1/2)$ .

Q.E.D.