Chapter 2.3, Problem 83E

Solution

a.

To confirm that the point $Q(2,7)$ is a point of intersection of the graph of the function $f\left(x\right)=-x^3+2x^2+9x-11$ and that of the line $y=5(x-2)+7$ .

Given:

The function:

  $f\left(x\right)=-x^3+2x^2+9x-11$

The line:

  $y=5(x-2)+7$

Confirmation:

Confirming that $Q(2,7)$ is a point on the line $y=5(x-2)+7$ :

Substitute $x=2$ in the equation of the line $y=5(x-2)+7$ .

  $\begin{array}{l}y=5(2-2)+7\\ y=5\cdot 0+7\\ y=7\end{array}$ .

Thus, the equation of the line $y=5(x-2)+7$ gives $y=7$ when it is substitute with $x=2$ .

That is, $Q(2,7)$ is a point on the line $y=5(x-2)+7$ .

Confirming that $Q(2,7)$ is a point on the graph of the function $f\left(x\right)=-x^3+2x^2+9x-11$ :

Substitute $x=2$ in the equation $f\left(x\right)=-x^3+2x^2+9x-11$ .

  $\begin{array}{l}f\left(x\right)=-2^3+2\cdot 2^2+9\cdot 2-11\\ f\left(x\right)=-8+8+18-11\\ f\left(x\right)=7\end{array}$ .

Thus, $f\left(x\right)=7$ when $x=2$ .

That is, $Q(2,7)$ is a point on the graph of the function also.

Now, since the point $Q(2,7)$ belongs to both the graphs, $Q(2,7)$ is a point on the intersection of the graphs.

b.

To zoom in at the point $Q(2,7)$ to develop an understanding that $y=5(x-2)+7$ is a linear approximation of $f\left(x\right)=-x^3+2x^2+9x-11$ at $x=2$ .

The portion of the graph around $Q(2,7)$ is closely observed.

  

It is observed that $y=5(x-2)+7$ coincides with $f\left(x\right)=-x^3+2x^2+9x-11$ at the point $Q(2,7)$ .

Hence, it is concluded that $y=5(x-2)+7$ is a linear approximation of $f\left(x\right)=-x^3+2x^2+9x-11$ at $x=2$ .

c.

To view the graphs of $y=5(x-2)+7$ and $f\left(x\right)=-x^3+2x^2+9x-11$ in a $[-5,5]$ by $[-25,25]$ window and to explain why the definition of tangent of a circle is not applicable in this case.

It seems like the line $y=5(x-2)+7$ touches $f\left(x\right)=-x^3+2x^2+9x-11$ at infinitely many points. So, the definition of the tangents in the case of circles is inappropriate to use here.

The line $y=5(x-2)+7$ and the function $f\left(x\right)=-x^3+2x^2+9x-11$ is graphed in a $[-5,5]$ by $[-25,25]$ window:

  

It seems like the line $y=5(x-2)+7$ touches $f\left(x\right)=-x^3+2x^2+9x-11$ at infinitely many points.

So, the definition of the tangent in the case of circles is inappropriate to use here. The only possible way to define a tangent is to approximate such a line.