Q. 50 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$ to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility

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

Step-by-Step Solution

Answer

There is no Inflection point . The function is concave up on$(-\infty ,0)\text{ and concave down on }(0,\infty )\text{. }$

Step 1. Given information.

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

Step 2. Second Derivative.

On differentiating the function, we get,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\frac{2^x}{1-2^x} \\ =\frac{\left(2^x\ln 2\right)\left(1-2^x\right)-\left(2^x\right)\left(\frac{d}{dx}1-\frac{d}{dx}{2}^x\right)}{{\left(1-2^x\right)}^2} \\ =\frac{2^x\ln 2}{{\left(2^x-1\right)}^2} \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{2^x\ln 2}{{\left(2^x-1\right)}^2} \\ =-\frac{{\ln }^2\left(2\right)2^x\left(2^x+1\right)}{{\left(2^x-1\right)}^3} \end{gathered}$$

Step 3. Sign chart.

Now,

$f\text{'}\text{'}\left(x\right)=0 at x=0.$

Therefore,

the sign chart will be,

The function is$\text{ concave up on }(-\infty ,0)\text{ and concave down on }(0,\infty )$ and no inflection point as domain is $(-\infty ,0)\cup (0,\infty ).$

Step 4. Verification.

The graph of the function is,