Chapter 6.3, Problem 31E

To determine

Tofind: Whetherfunction $q$ will always have greater $y-$ value then function $p$ after $x=2$ .

Answer to Problem 31E

No, function $q$ will not always have greater $y-$ value then function $p$ after $x=2$ .

Explanation of Solution

Given information :

Function $p$ is an exponential function and function $q$ is a linear function.

Concept used:

Use patterns between consecutive data pairs to determine type of function.

Linear function: - The difference of consecutive y-values is constant.

Exponential function: - Consecutive y-values have a common ratio. In each case, the differences of consecutive x-values need to be constant.

  

  $\begin{array}{l}q(x)=mx+n\\ p(x)=a{b}^x\end{array}$

The two functions intersect in $(2,9)$ . We have,

  $2m+n=a{b}^2=9$

The graphs of the functions have either no point of intersection, one point of intersection (if they are tangent) or two points of intersection. Since the graphs show one point of intersection and no tangency, it means there is a second point of intersection.

As exponential growth functions like $p(x)$ will eventually overtake linear growth functions, it means $q(x)>p(x)$ only for $x\in (2,\alpha )$ , and then $p(x)>q(x)$ for $x>\alpha$ .

So, friend is wrong.

Hence,

No, function $q$ will not always have greater $y-$ value then function $p$ after $x=2$ .