Q. 77 - 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. 

$f\left(x\right)=\frac{e^x}{x}$

Step-by-Step Solution

Answer

The sign chart is  

The sketch of the graph is  

Step 1. Given Information.

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

Step 2. Finding the roots.

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

So,

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

Therefore, the given function is undefined at $x=0.$

Step 3. Testing the signs.

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

Let's differentiate the equation to find $f\text{'}.$

So, 

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

Thus, $f\text{'}$ has a local minimum at $x=1.$ It is positive on the interval $\left(1,\infty \right)$ and negative elsewhere. Hence the graph of f will be increasing during the positive interval and decrease during the negative interval.

Let's differentiate again.

So, 

$f\text{'}\text{'}\left(x\right)=\frac{\left(x^2-2x+2\right)e^x}{x^3}$

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

Step 4. Sketch the sign chart.

The sign chart is 

Step 5. Examine the relevant limit.

Let's examine the limits.

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

Step 6. Sketch the graph of function f.

The graph of the function is