Q. 76 - Calculus

Question

Sketch careful, labeled graphs of each function $f$ in Exercises $57-82$ by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f$ and $f^{\text{'}}$and examine any relevant limits so that you can describe all key points and behaviors of $f$.

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

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=\frac{2^x}{1-2^x}$ is,

Step 1 . Given information

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

Step 2 . Let y = 2 x 1 - 2 x .

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-2$	$0.3333$	$\left(-2,0.3333\right)$
$-1$	$1$	$\left(-1,1\right)$
$1$	$-2$	$\left(1,-2\right)$
$4$	$-1.3333$	$\left(4,-1.3333\right)$

Step 3 . The graph for the function is,

Step 4 . Now for critical point f ' x = 0 .

$$\begin{gathered} \frac{d}{dx}\left(\frac{2^x}{1-2^x}\right)=0 \\ \frac{\left(1-2^x\right)2^x\ln 2-2^x\left(-2^x\ln 2\right)}{{\left(1-2^x\right)}^2}=0 \\ \left(1-2^x\right)2^x\ln 2-2^x·2^x\ln 2=0 \\ {2}^x\ln 2\left[1-2^x-2^x\right]=0 \\ {2}^x\ln 2\left[1-2·2^x\right]=0 \\ {2}^x\ln 2\left[1-2^{x+1}\right]=0 \\ 1-2^{x+1}=0 \\ {2}^{x+1}=1 \\ x+1=0 \\ x=-1 \end{gathered}$$

Therefore, $f$ has a critical point at $x=-1$. It does not have a local extrema.

Step 5 . The sign chart of f is shown below:

For roots of the function,

$$\begin{gathered} \frac{2^x}{1-2^x}=0 \\ {2}^x=0 \end{gathered}$$

Therefore, the function has no roots.

Again,

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{2^x}{1-2^x} \\ =0 \\ \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\frac{2^x}{1-2^x} \\ =0 \end{gathered}$$

Therefore, the function is defined everywhere except at $x=0$.The function is positive on $\left(-\infty ,0\right)$ and negative on $\left(0,\infty \right)$.The function is increasing on $\left(-\infty ,0\right)$ and the limits are $\underset{x\to \infty }{\lim }f\left(x\right)=0; \underset{x\to -\infty }{\lim }f\left(x\right)=0$