Chapter 12.2, Problem 46E

a.

To determine

To find: To sketch the graph of the given function.

Answer to Problem 46E

  

The above graph is the sketch for each piecewise function.

Explanation of Solution

Given:$C\left(x\right)=\{\begin{array}{l}0.015x,0<x\le 25\\ 0.10x,25<x\le 100\\ 0.07x,100<x\le 500\\ 0.05x,x>500\end{array}$

  

Thus the given piecewise functions can be graphed.

b.

To determine

To find: To find each limit and interpret the result in the context of the situation.

Answer to Problem 46E

1. $\underset{x\to 15}{\lim }C\left(x\right)$ gives the cost of making $15$ photocopies is $\$2.25$ .
2. $\underset{x\to 99}{\lim }C\left(x\right)$ gives the cost of making $99$ photocopies is $\$9.90$ .
3. $\underset{x\to 305}{\lim }C\left(x\right)$ gives the cost of making $305$ photocopies is $\$21.35$ .

Explanation of Solution

Given:$C\left(x\right)=\{\begin{array}{l}0.015x,0<x\le 25\\ 0.10x,25<x\le 100\\ 0.07x,100<x\le 500\\ 0.05x,x>500\end{array}$

  

1. $\underset{x\to 15}{\lim }C\left(x\right)=0.15\left(15\right)=2.25$

Gives the cost of making $15$ photocopies is $\$2.25$ .

2. $\underset{x\to 99}{\lim }C\left(x\right)=0.10\left(99\right)=9.9$

Gives the cost of making $99$ photocopies is $\$9.90$ .

3. $\underset{x\to 305}{\lim }C\left(x\right)=0.07\left(305\right)=21.35$

Gives the cost of making $305$ photocopies is $\$21.35$ .

Thus we can find the each limit and interpret the result in the context of the situation.

c.

To determine

To find: To create a table of values to show numerically that each limit does not exist.

Answer to Problem 46E

1. $\underset{x\to 25}{\lim }C\left(x\right)$ does not exist.

x	22	23	24	25
C(x)	3.30	3.45	3.60	?

x	26	27	28
C(x)	2.60	2.70	2.80

2. $\underset{x\to 100}{\lim }C\left(x\right)$ does not exist.

x	97	98	99	100
C(x)	9.70	9.80	9.90	?

x	101	102	103
C(x)	7.07	7.14	7.21

3. $\underset{x\to 500}{\lim }C\left(x\right)$ does not exist.

x	497	498	499	500
C(x)	34.79	34.86	34.93	?

x	501	502	503
C(x)	25.05	25.10	25.15

Explanation of Solution

Given:$C\left(x\right)=\{\begin{array}{l}0.015x,0<x\le 25\\ 0.10x,25<x\le 100\\ 0.07x,100<x\le 500\\ 0.05x,x>500\end{array}$

  

For $\underset{x\to {25}^{-}}{\lim }C\left(x\right)$ use $0.15x$ , $\underset{x\to {25}^{+}}{\lim }C\left(x\right)$ use $0.10x$ .

x	22	23	24	25
C(x)	3.30	3.45	3.60	?

x	26	27	28
C(x)	2.60	2.70	2.80

From the above table we can understand that $\underset{x\to {25}^{-}}{\lim }C\left(x\right)$ is not equal to $\underset{x\to {25}^{+}}{\lim }C\left(x\right)$ .

Therefore the limit $\underset{x\to 25}{\lim }C\left(x\right)$ does not exist.

For $\underset{x\to {100}^{-}}{\lim }C\left(x\right)$ use $0.10x$ , $\underset{x\to {25}^{+}}{\lim }C\left(x\right)$ use $0.07x$ .

x	97	98	99	100
C(x)	9.70	9.80	9.90	?

x	101	102	103
C(x)	7.07	7.14	7.21

From the above table we can understand that $\underset{x\to {100}^{-}}{\lim }C\left(x\right)$ is not equal to $\underset{x\to {100}^{+}}{\lim }C\left(x\right)$ .

Therefore the limit $\underset{x\to 100}{\lim }C\left(x\right)$ does not exist.

For $\underset{x\to {500}^{-}}{\lim }C\left(x\right)$ use $0.07xx$ , $\underset{x\to {500}^{+}}{\lim }C\left(x\right)$ use $0.05x$ .

x	497	498	499	500
C(x)	34.79	34.86	34.93	?

x	501	502	503
C(x)	25.05	25.10	25.15

From the above table we can understand that $\underset{x\to {500}^{-}}{\lim }C\left(x\right)$ is not equal to $\underset{x\to {500}^{+}}{\lim }C\left(x\right)$ .

Therefore the limit $\underset{x\to 500}{\lim }C\left(x\right)$ does not exist.

Thus we can create a table of values to show numerically that each limit does not exist.

d.

To determine

To find: To explain how to use the graph in part ‘a’ to verify that the limits in part ‘c’ do not exist.

Answer to Problem 46E

In part ‘c’ the function $C\left(x\right)$ approaches a different valuef from left to right of the domain values.

Explanation of Solution

Given:$C\left(x\right)=\{\begin{array}{l}0.015x,0<x\le 25\\ 0.10x,25<x\le 100\\ 0.07x,100<x\le 500\\ 0.05x,x>500\end{array}$

  

Each value of the input in part ‘c’ has a hole and a solid dot which indicates the function $C\left(x\right)$ approaches a different value from the left and right of the domains of ‘c’.

Thus we can verify the limits in part ‘c’ do not exist.