Q. 79 - 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)=x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Step-by-Step Solution

Answer

The sketch of the graph is  

The sign chart is

Step 1. Given Information.

The given function is $f\left(x\right)=x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

Step 2. Finding the roots.

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

So,

$$\begin{gathered} f\left(x\right)=x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ 0=x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ 0=x(x-1) \\ x=0 and x-1=0 \\ x=1 \end{gathered}$$

Therefore, the given function has roots at $x=0,1.$

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, 

  $$\begin{gathered} f\text{'}\left(x\right)=\frac{2}{3x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ 0=\frac{2}{3x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ x=\frac{1}{8} \end{gathered}$$

Thus, $f\text{'}$ is positive on the interval $\left(\frac{1}{8},\infty \right)$ 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, 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{2}{9x^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{2}{9x^{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ 0=\frac{2}{9x^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{2}{9x^{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ x=1 \end{gathered}$$

Thus, $f\text{'}\text{'}$ has inflection point at $x=1.$ It is positive on the interval $\left(-\infty ,1\right)$ and negative on the interval $\left(1,\infty \right).$ Hence, the graph of f will be concave up on the positive interval and concave down on the negative interval.

Step 4. Sketch the sign chart.

The sign chart is 

Step 5. Examine the relevant limit.

Let's examine the limits.

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

Step 6. Sketch the graph of function f.

The graph of the function is