Chapter 1.2, Problem 71E

(a)

To determine

To draw: Thegraph of $y_3$ in the interval $\left[-3,3\right]$ by $\left[-1,3\right]$ .

Explanation of Solution

Given information:The given functions are $y_1=\sqrt{x}$ and $y_2=\sqrt{1-x}$ . Also, $y_3=y_1+y_2$ .

Graph:

To draw the graph of $y_3$ by the graphing calculator, follow the steps given below.

Step 1: Press “ON” button on graphical calculator and press $\boxed{\text{Y}=}$ key to graph the functions.

Step 2: Enter the keystrokes $\boxed{2\text{nd}}\boxed{x^2}\boxed{\text{X}}\boxed{)}$ after the symbol ${\text{Y}}_{\text{1}}$ for the equation $y_1=\sqrt{x}$ . Now, press $\boxed{\text{ENTER}}$ and enter the keystrokes $\boxed{2\text{nd}}\boxed{x^2}\boxed{1}\boxed{-}\boxed{\text{X}}\boxed{)}$ after the symbol ${\text{Y}}_{\text{2}}$ for the equation $y_2=\sqrt{1-x}$ .

Step 3: Turn off ${\text{Y}}_{\text{1}}$ and ${\text{Y}}_{\text{2}}$ by un-highlighting their equal signs.

Step 4: Now, to write the equation $y_3=y_1+y_2$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}\boxed{+}\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}$ .

Step 5: Set the window at the interval $\left[-3,3\right]$ by $\left[-1,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (1)

Interpretation: From the graph it can be observed that the domain of the function $y_3$ is $\left[0,1\right]$ .

(b)

To determine

To find: The domain of $y_1$ , $y_2$ and $y_3$ .

Answer to Problem 71E

The domain of $y_1$ is $\left[0,\infty \right)$ , $y_2$ is $\left(-\infty ,1\right]$ and $y_3$ is $\left[0,1\right]$ .

Explanation of Solution

Given information:The given functions are $y_1=\sqrt{x}$ and $y_2=\sqrt{1-x}$ . Also, $y_3=y_1+y_2$ .

Calculation:

The domain of a function is the set of all $x$ values that define the function.

The function $y_1=\sqrt{x}$ is defined for all non-negative values of x . As negative values of x does not give real values. So, its domain is $\left[0,\infty \right)$ .

The function $y_2=\sqrt{1-x}$ is defined for all values of x less than 1. As the values of x greater than 1 does not give real value. So, its domain is $\left(-\infty ,1\right]$ .

From part (a), the domain of the function $y_3$ is $\left[0,1\right]$ .

Therefore, the domain of $y_1$ is $\left[0,\infty \right)$ , $y_2$ is $\left(-\infty ,1\right]$ and $y_3$ is $\left[0,1\right]$ .

(c)

To determine

To find: The domains of functions $y_3=y_1-y_2$ , $y_3=y_2-y_1$ , $y_3=y_1\cdot y_2$ , $y_3=\frac{y_1}{y_2}$ and $y_3=\frac{y_2}{y_1}$ .

Answer to Problem 71E

The domain of the functions $y_3=y_1-y_2$ , $y_3=y_2-y_1$ and $y_3=y_1\cdot y_2$ is the same $\left[0,1\right]$ . The domain of the function $y_3=\frac{y_1}{y_2}$ is $\left[0,1\right)$ and $y_3=\frac{y_2}{y_1}$ is $\left(0,1\right]$ .

Explanation of Solution

Given information:The given functions are $y_1=\sqrt{x}$ and $y_2=\sqrt{1-x}$ . Consider the functions $y_3=y_1-y_2$ , $y_3=y_2-y_1$ , $y_3=y_1\cdot y_2$ , $y_3=\frac{y_1}{y_2}$ and $y_3=\frac{y_2}{y_1}$ .

Calculation:

To draw the graph of $y_3=y_1-y_2$ by the graphing calculator, repeat the same steps from step 1 to step 3 as used in part (a) then follow the steps given below.

To write the equation $y_3=y_1-y_2$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}\boxed{-}\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}$ .

Now, set the window at the interval $\left[-3,3\right]$ by $\left[-3,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (2)

To draw the graph of $y_3=y_1-y_2$ by the graphing calculator, repeat the same steps from step 1 to step 3 as used in part (a) then follow the steps given below.

To write the equation $y_3=y_2-y_1$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}\boxed{-}\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}$ .

Now, set the window at the interval $\left[-3,3\right]$ by $\left[-3,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (3)

To write the equation $y_3=y_1\cdot y_2$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}\boxed{\times }\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}$ .

Now, set the window at the interval $\left[-3,3\right]$ by $\left[-3,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (4)

To write the equation $y_3=\frac{y_1}{y_2}$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}\boxed{÷}\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}$ .

Now, set the window at the interval $\left[-3,3\right]$ by $\left[-3,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (5)

To write the equation $y_3=\frac{y_2}{y_1}$ after the symbol ${\text{Y}}_{\text{3}}$ , press $\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{2}\boxed{\text{ENTER}}\boxed{÷}\boxed{\text{VARS}}\boxed{\Rightarrow }\boxed{1}\boxed{1}\boxed{\text{ENTER}}$ .

Now, set the window at the interval $\left[-3,3\right]$ by $\left[-3,3\right]$ and press $\boxed{\text{GRAPH}}\boxed{\text{TRACE}}$ key to produce the graph of function as shown below.

  

Figure (6)

Therefore, from the graph above graphs it can be interpreted that the domain of the functions $y_3=y_1-y_2$ , $y_3=y_2-y_1$ and $y_3=y_1\cdot y_2$ is the same $\left[0,1\right]$ . The domain of the function $y_3=\frac{y_1}{y_2}$ is $\left[0,1\right)$ and $y_3=\frac{y_2}{y_1}$ is $\left(0,1\right]$ .

(d)

To determine

To find: The conjecture that can be made about the domain of sums, difference, products and quotients of functions.

Answer to Problem 71E

The domain of sum, difference and products of two functions is the intersection of the domains of given functions.The domain of quotient of functions is the intersection of their domains by removal of any zeros of the denominators.

Explanation of Solution

Given information:The given functions are $y_1=\sqrt{x}$ and $y_2=\sqrt{1-x}$ .

The domain of the function $y_1=\sqrt{x}$ $\left[0,\infty \right)$ and the function $y_2=\sqrt{1-x}$ is $\left(-\infty ,1\right]$ .

From part (a), the domain of the function $y_3=y_1+y_2$ is $\left[0,1\right]$ .

From part (b), the domain of the functions $y_3=y_1-y_2$ , $y_3=y_2-y_1$ and $y_3=y_1\cdot y_2$ is the same $\left[0,1\right]$ . The domain of the function $y_3=\frac{y_1}{y_2}$ is $\left[0,1\right)$ and $y_3=\frac{y_2}{y_1}$ is $\left(0,1\right]$ .

From part (a) and part (b), it can be observed that the domain of sum, difference and products of two functions is the intersection of the domains of given functions.

The domain of quotient of functions is the intersection of their domains by removal of any zeros of the denominators.