Chapter 2.3, Problem 20E

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: $f\left(x\right)=x^3-2x^2-41x+42$

Graph:

Sketch the graph of $f\left(x\right)=x^3-2x^2-41x+42$with all its extrema and zeros.

  

Find the extrema:

Find the first derivative of the function:

  $f\left(x\right)=3x^2-4x-41$

Find the second derivative of the function:

  $f^{\prime \prime }\left(x\right)=6x-4$

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

  $3x^2-4x-41=0$

Use the quadratic formula to find the solutions:

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

Substitute the values $a=3,b=-4,$ and $c=-41$ into the quadratic formula and solve for $x$ :

  $x=\frac{4\pm \sqrt{{\left(-4\right)}^2-4\cdot \left(3\times \left(-41\right)\right)}}{2\cdot 3}$

The values of $x$ are $x=\left(\frac{2+\sqrt{127}}{3},\frac{2-\sqrt{127}}{3}\right)$

Find the $y-$ value:

Replace the variable $x$ with $x=\frac{2+\sqrt{127}}{3}$ in the expression $f\left(x\right)=x^3-2x^2-41x+42$ :

  $f\left(\frac{2+\sqrt{127}}{3}\right)={\left(\frac{2+\sqrt{127}}{3}\right)}^3-2{\left(\frac{2+\sqrt{127}}{3}\right)}^2-41\frac{2+\sqrt{127}}{3}+42$

Simplify the result:

  $y=\frac{380-254\sqrt{127}}{27}$

So, the $y-$ values are $\left(\frac{380-254\sqrt{127}}{27},\frac{380+254\sqrt{127}}{27}\right)$

The local minima:

  $\begin{array}{l}x=\frac{2+\sqrt{127}}{3}\text{or}4.423\\ y=\frac{380-254\sqrt{127}}{27}\text{or}-91.942\end{array}$

The local maxima:

  $\begin{array}{l}x=\frac{2-\sqrt{127}}{3}\text{or}-3.089\\ y=\frac{380+254\sqrt{127}}{27}\text{or}120.090\end{array}$

The function $f$ has a local maximum of about $120.090$ at $x\approx -3.089$ , and a local minimum of about $-91.942$ at $x\approx 4.423$ .

Find the zeroes of the function $f\left(x\right)=x^3-2x^2-41x+42$:

Factor $x^3-2x^2-41x+42$ :

  $\left(x-1\right)\left(x^2-x-42\right)=0$

Factor $x^2-x-42$ :

  $\left(x-1\right)\left(x-7\right)\left(x+6\right)=0$

Find the values of $x:$

  $\begin{array}{c}x-1=0\\ x-7=0\\ x+6=0\end{array}$

The function has zeros at $x=-6,x=1,x=7.$

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.