Chapter 1.3, Problem 20E

(a)

To determine

To find: The domain of the graph.

Answer to Problem 20E

Domain is: R

Explanation of Solution

Given data:

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

Calculation:

Domain of function:

All real numbers as the function is real for every real number.

Domain is: R (real numbers).

Conclusion:

Domain is:$\text{(-R,R)}$

(b)

To determine

To find: The range of the graph.

Answer to Problem 20E

The range of function is :$(4,-\infty )$

Explanation of Solution

Given data:

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

Calculation:

Range of function:

As the graph is increased till 4 units and then starts decreasing so 4 unit is the peak of the graph .

The range of function is :$(4,-\infty )$

Conclusion:

The range of function is :$(4,-\infty )$:$(4,-\infty )$

(c)

To determine

To find:The value of x at f(x)=0

Answer to Problem 20E

At f(x)=0 ,x=±2

Explanation of Solution

Given data:

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

Calculation:

Value of x for which f(x)=0

As the function f(x)=0 at the $4-x^2$So,

  $\begin{array}{l}4-x^2=0\\ x=\pm 2\end{array}$

Conclusion:

At f(x)=0 ,x=±2

(d)

To determine

To find: value of f(0)

Answer to Problem 20E

The value of f(0)=4

Explanation of Solution

Given data:

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

Calculation:

The value of f(0)

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

Conclusion:

The value of f(0)=4

(e)

To determine

To find:Coordinates of f(1).

Answer to Problem 20E

f(1)=3 so coordinates are (1,3)

Explanation of Solution

Given data:

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

Calculation:

For x =1 we have to take $f(x)=4-x^2$

So,

  $\begin{array}{l}f(x)=4-x^2\\ f(1)=4-1\\ f(1)=3\end{array}$

Coordinates are (1,3)

Conclusion:

f (1)=3 so coordinates are (1,3)

(f)

To determine

To find:Coordinates of f (-3)

Answer to Problem 20E

f (-3)=1 so the coordinates are (-3,1)

Explanation of Solution

Given data:

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

Calculation:

For value of x <0

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

So, the coordinates are (-3,1)

Conclusion:

f (-3)=1 so the coordinates are (-3,1)