Chapter 2.3, Problem 17E

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: $f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(x+3\right)$

Graph:

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

  

Find the extrema:

Find the first derivative of the function.

  $f\left(x\right)=3x^2+8x+1$

Find the second derivative of the function.

  $f^{\prime \prime }(x)=6x+8$

To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.

  $3x^2+8x+1=0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a=3,b=8,$ and $c=1$ into the quadratic formula and solve for $x$ .

  $\begin{array}{l}x=\frac{-8\pm \sqrt{8^2-4\cdot \left(3\cdot 1\right)}}{2\cdot 3}\\ x=\frac{-4\pm \sqrt{13}}{3}\end{array}$

The values of $x$ are $x=\frac{-4+\sqrt{13}}{3}$ and $x=\frac{-4-\sqrt{13}}{3}$

Find the $y-$ value:

Replace the variable $x$ with $-\frac{4-\sqrt{13}}{3}$ in the expression.

  $f\left(-\frac{4-\sqrt{13}}{3}\right)=\left(\left(-\frac{4-\sqrt{13}}{3}\right)-1\right)\left(\left(-\frac{4-\sqrt{13}}{3}\right)+2\right)\left(\left(-\frac{4-\sqrt{13}}{3}\right)+3\right)$

Simplify the result:

  $y=-\frac{70+26\sqrt{13}}{27}$

So, the $y-$ values are $y=-\frac{70+26\sqrt{13}}{27}$ and $y=-\frac{70-26\sqrt{13}}{27}$

The local minima:

  $\begin{array}{l}x=\frac{-4+\sqrt{13}}{3}\text{or}-0.131,\\ y=-\frac{70+26\sqrt{13}}{27}\text{or}-6.064\end{array}$

The local maxima:

  $\begin{array}{l}x=\frac{-4-\sqrt{13}}{3}\text{or}-2.535\\ y=-\frac{70-26\sqrt{13}}{27}\text{or}0.879\end{array}$

The function $f$ has a local maximum of about $0.879$ at $x\approx -2.535$ , and a local minimum of about $-6.064$ at $x\approx -0.131$ .

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

  $\begin{array}{l}x-1=0\\ x+2=0\\ x+3=0\end{array}$

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

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