Question

How do you sketch the curve $f\left( x \right)=\dfrac{{{e}^{x}}}{1+{{e}^{x}}}$?

Answer 1

Now to find the graph of the curve we will find the first derivative of the function and check if the function is increasing or decreasing. Now we will again differentiate to find the second derivative and find if the function is concave upwards or concave downwards. Now we will also find the t intercept by substituting x = 0 and hence we can easily draw a graph of the given function.

Complete step by step solution:
Now let us consider the given function $f\left( x \right)=\dfrac{{{e}^{x}}}{1+{{e}^{x}}}$
First let us check the nature of the function.
Now we will check if the function is increasing or decreasing.
Now we know that if $f\left( x \right)=\dfrac{g}{h}$ then $f'\left( x \right)=\dfrac{hg'-gh'}{{{h}^{2}}}$
Hence using this we get,
$f'\left( x \right)=\dfrac{{{e}^{x}}\left( 1+{{e}^{x}} \right)-{{e}^{x}}\left( {{e}^{x}} \right)}{{{\left( 1+{{e}^{x}} \right)}^{2}}}$
Hence we get,

& \Rightarrow f'\left( x \right)=\dfrac{{{e}^{x}}\left( 1+{{e}^{x}}-{{e}^{x}} \right)}{{{\left( 1+{{e}^{x}} \right)}^{2}}} \\\ & \Rightarrow f'\left( x \right)=\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)} \\\ \end{aligned}$$ Hence we can say that $f'\left( x \right)>0$ for all x. Now let us check the second derivative of the function. Hence again differentiating the function we get, $\begin{aligned} & \Rightarrow f''\left( x \right)=\dfrac{{{e}^{x}}{{\left( 1+{{e}^{x}} \right)}^{2}}-2{{e}^{x}}\left( 1+{{e}^{x}} \right)}{{{\left( 1+{{e}^{x}} \right)}^{4}}} \\\ & \Rightarrow f''\left( x \right)=\dfrac{{{e}^{x}}\left( 1+{{e}^{x}} \right)\left[ 1+{{e}^{x}}-2 \right]}{{{\left( 1+{{e}^{x}} \right)}^{4}}} \\\ \end{aligned}$ $\Rightarrow f''\left( x \right)=\dfrac{{{e}^{x}}\left( {{e}^{x}}-1 \right)}{\left( 1+{{e}^{x}} \right)}$ Now we know that ${{e}^{x}}>1,x<0$ and ${{e}^{x}}<1,x>0$ Hence $f''\left( x \right)>0$ for $x\in \left( -\infty ,0 \right)$ and $f''\left( x \right)<0$ for $x\in \left( 0,\infty \right)$ Hence function is concave upwards for $x\in \left( -\infty ,0 \right)$ and the function is concave downwards for $x\in \left( 0,\infty \right)$ Now we on substituting x = 0 in the given function we get, $\Rightarrow f\left( x \right)=\dfrac{{{e}^{0}}}{1+{{e}^{0}}}=\dfrac{1}{2}$ Hence the point (0, 0.5) lies on the graph and is the y intercept of the graph. Hence now we have a general idea of the graph. Now plotting according to the obtained conditions we get the graphs as,  **Note:** Now note that here we got the function as purely increasing. It is possible that the function is increasing and decreasing depending on the intervals taken. Hence we can find these intervals in the first derivative test itself. Also note that if the first derivative is 0 then we get an extrema in the function.