Q. 70 - Calculus

Question

Sketch careful, labeled graphs of each function f in Exercises 63–82 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 $f, f\text{'}, and f\text{'}\text{'},$ and examine any relevant limits so that you can describe all key points and behaviors of f. 

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

Step-by-Step Solution

Answer

The sign chart is 

The sketch of the graph is 

Step 1. Given Information.

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

Step 2. Finding the roots and examining the relevant limit.

To find the roots we will put the given function equal to zero.

So,

 $$\begin{gathered} f\left(x\right)=\frac{x}{x^2+1} \\ 0=\frac{x}{x^2+1} \\ x=0 \end{gathered}$$

Let's examine the limits of $f\left(x\right)=\frac{x}{x^2+1} \mathrm{as} x\to \pm \infty .$

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

Step 3. Testing the signs.

To sketch the sign chart, let's differentiate the equation to find $f\text{'}.$

So,

 $$\begin{gathered} f\text{'}\left(x\right)=-\frac{x^2-1}{{\left(x^2+1\right)}^2} \\ 0=-\frac{x^2-1}{{\left(x^2+1\right)}^2} \\ 0=-x^2+1 \\ {x}^2=1 \\ x=\pm 1 \end{gathered}$$

Testing the signs on both sides,

$$\begin{gathered} f\text{'}(-2)=-\frac{{\left(-2\right)}^2-1}{{\left({\left(-2\right)}^2+1\right)}^2} \\ f\text{'}(-2)=-\frac{3}{25} \\ Now, f\text{'}\left(0\right)=-\frac{{\left(0\right)}^2-1}{{\left(0^2+1\right)}^2} \\ f\text{'}\left(0\right)=1 \\ Now, f\text{'}\left(2\right)=-\frac{{\left(2\right)}^2-1}{{\left(2^2+1\right)}^2} \\ f\text{'}\left(2\right)=-\frac{3}{25} \end{gathered}$$

Thus, $f\text{'}$ is positive on the interval $\left(-1,1\right)$ and negative on the interval $\left(-\infty ,-1\right) and \left(1,\infty \right).$ Hence the graph of f will be increasing on the positive intervals and decreasing on the negative intervals.

Step 4. Testing the signs.

Now, let's test the sign for $f\text{'}\text{'}.$

Let's differentiate again.

So, 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{2x\left(x^2-3\right)}{{\left(x^2+1\right)}^3} \\ 0=\frac{2x\left(x^2-3\right)}{{\left(x^2+1\right)}^3} \\ 0=2x\left(x^2-3\right) \\ x=0 and 0=x^2-3 \\ x^2=3 \\ x=\pm \sqrt{3} \end{gathered}$$

Thus, $f\text{'}\text{'}$ is positive on the intervals $\left(-\sqrt{3},0\right),\left(\sqrt{3},\infty \right)$ and negative on the intervals $\left(-\infty ,-\sqrt{3}\right),\left(0,\sqrt{3}\right).$ Hence the graph of f will be concave up on the positive intervals and concave down on the negative intervals.

Step 5. Sketch the sign chart.

The sign chart is 

Step 6. Sketch the graph of function f.

The graph of the function is