Chapter 2.3, Problem 21E

Solution

To Sketch: The given function with local and minimum extrema and $x-$ intercepts. And explain its end behaviour with upper and lower limits.

Given: $f\left(x\right)={\left(x-2\right)}^3\left(x+1\right)\left(x-3\right)$

Graph:

Sketch the graph of $f\left(x\right)={\left(x-2\right)}^3\left(x+1\right)\left(x-3\right)$with all its extrema and zeros.

  

Find the extrema:

Calculate the derivative of the function $f\left(x\right)={\left(x-2\right)}^3\left(x+1\right)\left(x-3\right)$ :

Apply the product rule ${\left[uvw\right]}^{\prime }=u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }$ :

  $\begin{array}{c}=\frac{d}{dx}\left[x-3\right]\cdot {\left(x-2\right)}^3\left(x+1\right)+\left(x-3\right)\cdot \frac{d}{dx}\left[{\left(x-2\right)}^3\right]\cdot \left(x+1\right)+\left(x-3\right){\left(x-2\right)}^3\cdot \frac{d}{dx}\left[x+1\right]\\ =\left(\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[-31\right]\right){\left(x-2\right)}^3\left(x+1\right)+\left(x-3\right)\cdot 3{\left(x-2\right)}^2\cdot \frac{d}{dx}\left[x-2\right]\cdot \left(x+1\right)+\left(x-3\right){\left(x-2\right)}^3\left(\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[1\right]\right)\\ =\left(1+0\right){\left(x-2\right)}^3\left(x+1\right)+\left(x-3\right)\cdot 3{\left(x-2\right)}^2\left(\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[-2\right]\right)\left(x+1\right)+\left(x-3\right){\left(x-2\right)}^3\left(1+0\right)\\ ={\left(x-2\right)}^3\left(x+1\right)+\left(x-3\right)\cdot 3{\left(x-2\right)}^2\left(1+0\right)\left(x+1\right)+\left(x-3\right){\left(x-2\right)}^3\\ ={\left(x-2\right)}^3\left(x+1\right)+3\left(x-3\right){\left(x-2\right)}^2\left(x+1\right)+\left(x-3\right){\left(x-2\right)}^3\end{array}$

Simplify the term:

  ${\left(x-2\right)}^2\left(5x^2-12x-5\right)$

Find the values of $x$ :

Set ${\left(x-2\right)}^2$ equal to $0$ :

  $\begin{array}{c}{\left(x-2\right)}^2=0\\ x-2=0\\ x=-2\end{array}$

Set $5x^2-12x-5$ equal to $0$ :

  $5x^2-12x-=0$

Use the quadratic formula to find the solutions.

  $\frac{-b\pm \sqrt{b^2-4\left(ac\right)}}{2a}$

Substitute the values $a=5,b=-12,$ and $c=-5$ into the quadratic formula:

  $\begin{array}{l}x=\frac{12\pm \sqrt{{\left(-12\right)}^2-4\cdot \left(5\cdot \left(-5\right)\right)}}{2\cdot 5}\\ x=\frac{6\pm \sqrt{61}}{5}\end{array}$

The values of $x$ are $x=\left(\frac{6+\sqrt{61}}{5},\frac{6-\sqrt{61}}{5}\right)$

Find the $y-$ value:

Replace the variable $x$ with $x=\frac{6+\sqrt{61}}{5}$ in the expression $f\left(x\right)={\left(x-2\right)}^3\left(x+1\right)\left(x-3\right)$to find $y$ :

  $y={\left(\frac{6+\sqrt{61}}{5}-2\right)}^3\left(\frac{6+\sqrt{61}}{5}+1\right)\left(\frac{6+\sqrt{61}}{5}-3\right)$

Simplify the result:

  $y=\frac{43546-5734\sqrt{61}}{3125}$

So, the $y-$ values are $\left(\frac{43546-5734\sqrt{61}}{3125},\frac{43546+5734\sqrt{61}}{3125}\right)$

The local minima:

  $\begin{array}{l}x=\frac{6+\sqrt{61}}{5}\text{or}2.762\\ y=\frac{43546-5734\sqrt{61}}{3125}\text{or}-0.396\end{array}$

The local maxima:

  $\begin{array}{l}x=\frac{6-\sqrt{61}}{5}\text{or}-0.362\\ y=\frac{43546+5734\sqrt{61}}{3125}\text{or}28.265\end{array}$

The function $f$ has a local maximum of about $28.265$ at $x\approx -0.362$ , and a local minimum of about $-0.396$ at $x\approx 2.762$ .

Find the zeroes of the function $f\left(x\right)={\left(x-2\right)}^3\left(x+1\right)\left(x-3\right)$:

  $\begin{array}{c}$ {(x-2)}^3=0$x+1=0\\ x-3=0\end{array}$

Find the values of $x:$

  $\begin{array}{c}x=2\\ x=-1\\ x=3\end{array}$

The function has zeros at $x=2,-1,3$

Thus, the function $f$ has $2$ extrema and $3$ zeros, the maximum number possible for a cubic.

End behaviour:

Use the limits to describe its end behaviour.

  $\underset{x\to \infty }{\lim }f\left(x\right)=\infty \text{ and }\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$

The graph Falls to the left and rises to the right.