Chapter 4.7, Problem 50E

a.

To determine

Describe how the graph of piecewise function changes when < is replaced with $\le$ and $\ge$ is replaced with > .Do the domain and range change.

Answer to Problem 50E

Domain is the same range changes

Explanation of Solution

Given information: $\begin{array}{l}f(x)=x+2,\text{if}x<2\text{and}\\ f(x)=-x-1\text{if}x\ge 2.\end{array}$

Calculation:

The domain of the original function $f(x)$ is all real numbers since combining $x<2andx\ge 2$ gives all real numbers.

Replacing < with $\le \text{and}\ge \text{with}>$ gives a new function,

  $\begin{array}{l}g(x)=x+2,\text{if}x\le 2\text{and}\\ g(x)=-x-1\text{if}x>2.\end{array}$

The domain of the function $g(x)$ is all real numbers since combining $x\le 2andx>2$ gives all real numbers.So,

The domains are then the same since the domain is all real numbers for both the new function and the original function.

For the original function ,the top equation of $y=x+2$ has a range of $y<4$ since $x<2\text{and}y=2+2=4$ .For the bottom equation of $y=-x-1$ has a range $y\le -3$ since $x\ge 2\text{and}y=-2-1=-3$ .Combining $y<4andy\le -3$ then gives a range $y<4$ for the original function.

For the newfunction ,the top equation of $y=x+2$ has a range of $y\le 4$ since $x\le 2\text{and}y=2+2=4$ .For the bottom equation of $y=-x-1$ has a range $y\le -3$ since $x\ge 2\text{and}y=-2-1=-3$ .Combining $y\le 4andy<-3$ then gives a range $y\le 4$ for the new function.

The range then changed since the original function has a range of $y<4$ and new function has a range of $y\le 4$ .

b.

To determine

Describe how the graph of piecewise function changes when < is replaced with $\le$ and $\ge$ is replaced with > .Do the domain and range change.

Answer to Problem 50E

Domain and range are the same.

Explanation of Solution

Given information: $\begin{array}{l}f(x)=(1/2)x+(3/2),\text{if}x<1\text{and}\\ f(x)=-x+3,\text{if}x\ge 1.\end{array}$

Calculation:

The domain of the original function $f(x)$ is all real numbers since combining $x<1andx\ge 1$ gives all real numbers.

Replacing < with $\le \text{and}\ge \text{with}>$ gives a new function,

  $\begin{array}{l}g(x)=(1/2)x+(3/2),\text{if}x\le 1\text{and}\\ g(x)=-x+3,\text{if}x>1.\end{array}$

The domain of the function $g(x)$ is all real numbers since combining $x\le 1\text{and}x>1$ gives all real numbers.So,

The domains are then the same since the domain is all real numbers for both the new function and the original function.

For the original function ,the top equation of $y=(1/2)x+(3/2)$ has a range of $y<2$ since $x<1\text{and}y=(1/2)\times 1+(3/2)=2$ .For the bottom equation of $y=-x+3$ has a range $y\le 2$ since $x\ge 1\text{and}y=-1+3=2$ .Combining $y<2andy\le 2$ then gives a range $y\le 2$ for the original function.

For the new function ,the top equation of $y=(1/2)x+(3/2)$ has a range of $y\le 2$ since $x\le 1\text{and}y=(1/2)\times 1+(3/2)=2$ .For the bottom equation of $y=-x+3$ has a range $y<2$ since $x>1\text{and}y=-1+3=2$ .Combining $y<2andy\le 2$ then gives a range $y\le 2$ for the new function.

The range are then the same since the original function has a range of $y\le 2$ and new function has a range of $y\le 2$ .