Q. 67 - 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. 

$f\left(x\right)=x^3-2x^2+x$

Step-by-Step Solution

Answer

The sign chart is 

The sketch of the graph is  

Step 1. Given Information.

The given function is $f\left(x\right)=x^3-2x^2+x.$

Step 2. Finding the roots.

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

So,

$$\begin{gathered} f\left(x\right)=x^3-2x^2+x \\ 0=x^3-2x^2+x \\ 0=x\left(x^2-2x+1\right) \\ 0=x\left(x-1\right)\left(x-1\right) \\ x=0 \mathrm{and} x-1=0 \\ x=1 \end{gathered}$$

Therefore, the given function have roots at $x=0 \mathrm{and} 1.$

Step 3. Testing the signs of f.

To sketch the sign chart, let's test the signs on both sides.

For f

$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3-2{\left(-1\right)}^2+\left(-1\right) \\ f\left(-1\right)=-4 \\ Now, f(0.5)={\left(0.5\right)}^3-2{\left(0.5\right)}^2+\left(0.5\right) \\ f(0.5)=0.125 \\ Now, f\left(2\right)={\left(2\right)}^3-2{\left(2\right)}^2+\left(2\right) \\ f\left(2\right)=2 \end{gathered}$$

Step 4. 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, 

 $$\begin{gathered} f\text{'}\left(x\right)=3x^2-4x+1 \\ 0=3x^2-4x+1 \\ 0=\left(3x-1\right)\left(x-1\right) \\ 3x-1=0 and x-1=0 \\ x=\frac{1}{3} and x=1 \end{gathered}$$

Testing the signs on both sides,

$$\begin{gathered} f\text{'}\left(0\right)=3{\left(0\right)}^2-4\left(0\right)+1 \\ f\text{'}\left(0\right)=1 \\ \mathrm{Now}, f\text{'}(0.5)=3{\left(0.5\right)}^2-4\left(0.5\right)+1 \\ f\text{'}(0.5)=-0.25 \\ \mathrm{Now}, f\text{'}\left(2\right)=3{\left(2\right)}^2-4\left(2\right)+1 \\ f\text{'}\left(2\right)=5 \end{gathered}$$

Thus, $f\text{'}$ is negative on the interval $\left(\frac{1}{3},1\right)$ and positive on the intervals $\left(-\infty ,\frac{1}{3}\right) and \left(1,\infty \right).$ Hence the graph of f will be increasing on the positive intervals and decreasing on the negative intervals.

Let's differentiate again.

So, 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=6x-4 \\ 0=2(3x-2) \\ 0\ne 2 and 0=3x-2 \\ x=\frac{2}{3} \end{gathered}$$

Thus, $f\text{'}\text{'}$ is positive on the interval . Hence, the graph of f will be concave up everywhere.

Step 5. Sketch the sign chart.

The sign chart is

 

Step 6. Examine the relevant limit.

Let's examine the limits of $f\left(x\right)=x^3-2x^2+x \mathrm{as} \mathrm{x}\to \pm \infty .$

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

Step 7. Sketch the graph of function f.

The graph of the function is