Chapter 2.3, Problem 19E

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: $f\left(x\right)=-x^3+4x^2+31x-70$

Graph:

Sketch the graph of $f\left(x\right)=-x^3+4x^2+31x-70$with all its extrema and zeros.

  

Find the extrema:

Find the first derivative of the function:

  $f\left(x\right)=-3x^2+8x+31$

Find the second derivative of the function:

  $f^{\prime \prime }\left(x\right)=-6x+8$

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

  $-3x^2+8x+31=0$

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 $c=31$ into the quadratic formula and solve for $x$ :

  $x=\frac{-8\pm \sqrt{8^2-4\cdot \left(-3\times \left(31\right)\right)}}{\left(2\cdot \left(-3\right)\right)}$

The values of $x$ are $x=\frac{4+\sqrt{109}}{3}$ and $x=\frac{4-\sqrt{109}}{3}$

Find the $y-$ value:

Replace the variable $x$ with $x=\frac{4+\sqrt{109}}{3}$ in the expression $f\left(x\right)=-x^3+4x^2+31x-70$ :

  $f\left(\frac{4+\sqrt{109}}{3}\right)=-{\left(\frac{4+\sqrt{109}}{3}\right)}^3+4{\left(\frac{4+\sqrt{109}}{3}\right)}^2+31\left(\frac{4+\sqrt{109}}{3}\right)-70$

Simplify the result:

  $y=\frac{125\sqrt{3}}{18}$

So, the $y-$ values are $y=\frac{125\sqrt{3}}{18}$ and $y=-\frac{125\sqrt{3}}{18}$

The local maxima:

  $\begin{array}{l}x=\frac{9+5\sqrt{3}}{6}\text{or}2.943,\\ y=\frac{125\sqrt{3}}{18}\text{or}12.028\end{array}$

The local minima:

  $\begin{array}{l}x=\frac{9-5\sqrt{3}}{6}\text{or}0.056\\ y=-\frac{125\sqrt{3}}{18}\text{or}-12.028\end{array}$

The function $f$ has a local maximum of about $12.028$ at $x\approx 2.943$ , and a local minimum of about $-12.028$ at $x\approx 0.056$ .

Find the zeroes of the function $f\left(x\right)=\left(2x-3\right)\left(4-x\right)\left(x+1\right)$:

Set $f\left(x\right)=\left(2x-3\right)\left(4-x\right)\left(x+1\right)$ equal to $0$ :

  $\begin{array}{c}2x-3=0\\ 4-x=0\\ x+1=0\end{array}$ or

Find the values of $x:$

  $\begin{array}{c}x=\frac{3}{2}\\ x=4\\ x=-1\end{array}$

The function has zeros at $x=\frac{3}{2},x=4,x=-1.$

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 rises to the left and falls to the right.