Chapter 1.3, Problem 18E

(a)

To determine

To Find: The domain of the function.

Answer to Problem 18E

Domain: all $x$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

Domain: all $x$

(b)

To determine

To Find: The range of the function.

Answer to Problem 18E

  $(-\infty ,4)$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

Range: $(-\infty ,4)$

(c)

To determine

To Find: The values for $x$ for which $f(x)=0$

Answer to Problem 18E

  $\begin{array}{l}x=-4\\ x=2\end{array}$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $f(x)=0$

  $\begin{array}{l}x=-4\\ x=2\end{array}$

(d)

To determine

To Find: The values from part $(c)$ referred to graphically?

Answer to Problem 18E

  $x$ Intercepts

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $\begin{array}{l}x=-4\\ x=2\end{array}$

They are $x$ intercepts.

(e)

To determine

To Find: $f(0)$ if possible.

Answer to Problem 18E

  $f(0)=4$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $\begin{array}{l}f(0)=0+4\\ f(0)=4\end{array}$

(f)

To determine

To Find: The values from part $(e)$ referred to graphically?

Answer to Problem 18E

  $y$ intercept

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $f(0)=4$ is a $y$ intercept.

(g)

To determine

To Find: The values of $f$ at $x=1$ ? What are the coordinates of the point?

Answer to Problem 18E

  $coordinate:(1,3)$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

For $f(1)$

  $\begin{array}{l}f(1)=4-1\\ f(1)=3\\ coordinate:(1,3)\end{array}$

(h)

To determine

To Find: The values of $f$ at $x=-1$ ? What are the coordinates of the point.

Answer to Problem 18E

  $coordinate:(-1,3)$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $\begin{array}{l}f(-1)=(-1)+4\\ f(-1)=3\\ coordinate:(-1,3)\end{array}$

(i)

To determine

To Find: The coordinates of the point on the graph at which $x=-3$ can be labeled $(-3,f(-3)),or(-3,?)$

Answer to Problem 18E

  $coordinate:(-3,1)$

Explanation of Solution

Given:

  

Given: $f(x)=\{\begin{array}{l}x+4,x\le 0\\ 4-x^2,x>0\end{array}$

  $\begin{array}{l}f(-3)=-3+4\\ f(-3)=1\\ coordinate:(-3,1)\end{array}$