Q. 44 - 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 

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

Step-by-Step Solution

Answer

The inflexion points are $x=\frac{1}{\sqrt{3}}\text{ and }x=-\frac{1}{\sqrt{3}}$and the function is concave up on$\left(-\infty ,-\frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}},\infty \right)$ and concave down in $\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).$

Step 1. Given Information.

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

Step 2. Second-derivative.

On differentiating,

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

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{-2x}{{\left(1+x^2\right)}^2} \\ =\frac{-2{\left(1+x^2\right)}^2+2x\left(2\left(1+x^2\right)2x\right)}{{\left(1+x^2\right)}^4} \\ =\frac{2\left(3x^2-1\right)}{{\left(1+x^2\right)}^3} \end{gathered}$$

Step 3. Sign Chart.

Now, $f\text{'}\text{'}\left(x\right)=0 at x=\frac{1}{\sqrt{3}}\text{ and }x=-\frac{1}{\sqrt{3}}$

Therefore, chart will be,

$$\begin{gathered} Concave up at\left(-\infty ,-\frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}},\infty \right) \\ \text{and concave down on }\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right) \\ \text{Inflection points:}x=\frac{1}{\sqrt{3}}\text{ and }x=-\frac{1}{\sqrt{3}} \end{gathered}$$

Step 4. Verification.

The graph of the function is,