Chapter 2.4, Problem 38AYU

In Problems 31-42:

(a) Find the domain of each function.

(b) Locate any intercepts.

(c) Graph each function.

(d) Based on the graph, find the range.

$f\left(x\right)=\{\begin{array}{l}2-x\text{if}-3\le x<1\\ \sqrt{x}\text{if}x>1\end{array}$

To determine

To find: The following values of the function, $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$:

a. Domain of the function $f$.

Answer to Problem 38AYU

Solution:

a. Domain of the function $f$: The real values which is greater than or equal to $-3$ excluding the number 1. That is, $\left[-3,1\right){{\displaystyle \cup }}^{\text{}}\left(1,\infty \right)$.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$

a. Domain of the function $f$: To find the domain of $f$, look at its definition. Since $f$ is defined for all $-3\le x<1$ and $x>1$, the domain of $f$ is That is, $\left[-3,1\right){{\displaystyle \cup }}^{\text{}}\left(1,\infty \right).$.

To determine

To find: The following values of the function, $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$:

b. Intercepts of the function $f$ if any.

Answer to Problem 38AYU

Solution:

b. Intercepts of the function $f$ if any: $y\text{-intercept is }(0,2)\text{ and there is no }x\text{- intercept for the given piecewise defined function}$.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$

b. Intercepts: The $y\text{-intercept}$ of the graph of the function is $f\left(0\right)$. Because the equation for $f$ when $x=0$ is $f\left(x\right)=2-x$, the $y\text{-intercept}$ is $y=2-0=2$.

The $x\text{- intercepts}$ of the graph of a function $f$ are the real solutions to the equation $f\left(x\right)=0$. To find the $x\text{- intercepts}$ of $f$, solve $f\left(x\right)=0$ for each “piece” of the function, and then determine which values of $x$, if any, satisfy the condition that defines the piece.

$f\left(x\right)=0$

$2-x=0$

$x=2$

Satisfies the condition $-3\le x<1$

$f\left(x\right)=0$

$\sqrt{x}=0$

$x=0$

Contradicts the condition $x>1$.

Therefore, there is no $x\text{- intercept}$.

To determine

To find: The following values of the function, $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$:

c. Graph of the function $f$.

Answer to Problem 38AYU

Solution:

c. Graph of the function $f$.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$

c.

To determine

To find: The following values of the function, $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$:

d. Range of the function $f$ based on its graph.

Answer to Problem 38AYU

Solution:

. Range of the function $f$ based on its graph: The range takes the positive real values which is greater than 1. $\text{Range}=\left(1,\infty \right)$.

Explanation of Solution

Given:

The function is: $f\left(x\right)=\{\begin{array}{c}2-x if -3\le x<1\\ \sqrt{x} if x>1\end{array}$

d. From the graph, it can be easily predicted that the range takes the positive real values which is greater than 1. $\text{Range}=\left(1,\infty \right)$