Chapter 0.2, Problem 71E

a.

To determine

To Graph: $y_3$in $\left[-3,3\right]$ by $\left[-1,3\right]$

Explanation of Solution

Given:

  $y_1=\sqrt{x}$

  $y_2=\sqrt{1-x}$ and

  $y_3=y_1+y_2$

Graph:

The graph of $y_3$in the given window in $\left[-3,3\right]$ by $\left[-1,3\right]$ will be

  

b.

To determine

To Compare: The domain of the graph of $y_3$ with the domains of the graphs of $y_1$ and $y_2$ .

Answer to Problem 71E

The domain of $y_3$ is the intersection of the domains of $y_1$ and $y_2$ .

Explanation of Solution

Given:

  $y_1=\sqrt{x}$

  $y_2=\sqrt{1-x}$ and

  $y_3=y_1+y_2$

Calculation:

The domain of $y_1=\sqrt{x}$ is $\left[0,\infty \right)$

The domain of $y_2=\sqrt{1-x}$ is $\left(-\infty ,1\right]$

The domain of $y_3=\sqrt{x}+\sqrt{1-x}$ is $\left[0,1\right]$

Comparison:

The domain of $y_1$ is $x\ge 0$ and the domain of $y_2$ is $x\le 1,$ then the domain of $y_3$ is $0\le x\le 1$ .

So, the domain of $y_3$ is the intersection of the domains of $y_1$ and $y_2$ .

c.

To determine

To Replace: $y_3$ by the given data's and compare it with part $\left(b\right)$ .

Answer to Problem 71E

The domain of $y_3$ intersects the domains of $y_2$ and $y_1$ in all cases. The domain of $y_3$ is same as the domains of $y_1-y_2,y_2-y_1$ and $y_1\cdot y_2$ , but varies in $\frac{y_1}{y_2}$ and $\frac{y_2}{y_1}$ for the extreme values.

Explanation of Solution

Given:

  $y_1=\sqrt{x}$

  $y_2=\sqrt{1-x}$ and

  $y_3=y_1+y_2$

Replace $y_3$ by $y_1-y_2,y_2-y_1,y_1\cdot y_2,\frac{y_1}{y_2}$ and $\frac{y_2}{y_1}$:

Calculation:

Replace $y_3$by $y_1-y_2$:

  $y_3=\sqrt{x}-\sqrt{1-x}$

The graph of $y_3=\sqrt{x}-\sqrt{1-x}$ is as follows:

  

Hence, the domain is $0\le x\le 1$ .

Replace $y_3$by $y_2-y_1$:

  $y_3=\sqrt{1-x}-\sqrt{x}$

The graph of $y_3=\sqrt{1-x}-\sqrt{x}$ is as follows:

  

Here the domain is $0\le x\le 1$ .

Replace $y_3$by $y_1\cdot y_2$:

  $y_3=\sqrt{x}\cdot \sqrt{1-x}$

The graph of $y_3=\sqrt{x}\cdot \sqrt{1-x}$ is as follows:

  

Hence, the domain is $0\le x\le 1$ .

Replace $y_3$by $\frac{y_1}{y_2}$:

  $y_3=\frac{\sqrt{x}}{\sqrt{1-x}}$

The graph of $y_3=\frac{\sqrt{x}}{\sqrt{1-x}}$ is as follows:

  

Here, the domain is $0\le x<1$ .

Replace $y_3$by $\frac{y_2}{y_1}$:

  $y_3=\frac{\sqrt{1-x}}{\sqrt{x}}$

The graph of $y_3=\frac{\sqrt{1-x}}{\sqrt{x}}$ is as follows:

  

Here the domain is $0<x\le 1$ .

Comparison with part $\left(b\right)$:

Thus, the domain of $y_3$ intersects the domains of $y_2$ and $y_1$ in all cases. The domain of $y_3$ is same as the domains of $y_1-y_2,y_2-y_1$ and $y_1\cdot y_2$ , but varies in $\frac{y_1}{y_2}$ and $\frac{y_2}{y_1}$ for the extreme values.

d.

To determine

To Conjecture: The domains of sums, differences, products, and quotients of functions based on the observations in $\left(b\right)$ and $\left(c\right)$ .

Answer to Problem 71E

Explanation of Solution

Given:

  $y_1=\sqrt{x}$

  $y_2=\sqrt{1-x}$ and

  $y_3=y_1+y_2$

Conjecture based on our results from $\left(a\right),\left(b\right)$and $\left(c\right)$:

While performing $y_1-y_2,y_2-y_1$ and $y_1\cdot y_2$ , if any of the resulting function values

that are not in the domains of original functions, that values should be excluded.

Thus, the resulting function's domain is the intersection of the original functions'

domains.

Also, while performing $\frac{y_1}{y_2}$ and $\frac{y_2}{y_1}$ if any of the resulting function values that are not in the domains of original functions, that values should be excluded. Along with that,

the domain will be excluded for values of the denominator where the function equals

zero.