Chapter 9.1, Problem 66PPS

(a)

To determine

To Write:

The domain of f(x)

Answer to Problem 66PPS

The domain is all real numbers.

Explanation of Solution

Given Information:

  $f(x)=x^2-9$

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Parent Function:

  $y=x^2$

• Standard Form :

  $f(x)=a{x}^2+bx+c;a\ne 0$

• Type of Graph:

Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.
• The domain of a parabola is all real numbers.
• The range of a parabola is all real numbers greater than and equal to the minimum value or all real numbers less than and equal to the maximum value.

Calculation:

  $f(x)=x^2-9$

Since it is a parabola ( quadratic equation) , so the domain of $f(x)=x^2-9$ is all real numbers.

(b)

To determine

To Find:

The range of f(x).

Answer to Problem 66PPS

Range = $\{y\in R:y\ge -9\}$

Where

R= set of all real numbers

Explanation of Solution

Given Information:

  $f(x)=x^2-9$

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Parent Function:

  $y=x^2$

• Standard Form :

  $f(x)=a{x}^2+bx+c;a\ne 0$

• Type of Graph:

Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.
• The domain of a parabola is all real numbers.
• The range of a parabola is all real numbers greater than and equal to the minimum value or all real numbers less than and equal to the maximum value.

Calculation:

In the equation $f(x)=x^2-9$; comparing it with standard form of quadratic equation , we have a = 1 , b = 0 , c = -9 .

Find equation of Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{0}{2(1)}=\mathrm{0..........................}[a=1,b=0]\end{array}$

The equation of axis of symmetry is $x=0$

Find the Vertex and determine whether it is maximum or minimum:

Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is $x=0$

Substituting the value of $x=0$in the equation $f(x)=x^2-9$

  $\begin{array}{l}f(x)=x^2-9\\ $ f(0)={(0)}^2-\mathrm{9...............}[x=0]$f(0)=0-\mathrm{9..................................}[simplify]\\ f(0)=-9\end{array}$

The vertex is at $(0,-9)$ .

Since the a = 1 > 0 .So, the graph opens upward .Hence, the vertex is the minimum.

So, the range is all real numbers greater than and equal to the minimum .

Here , the minimum value is -9.

Range = $\{y\in R:y\ge -9\}$

Where

R= set of all real numbers

(c)

To determine

To Find:

The values of x for which f(x) is negative.

Answer to Problem 66PPS

f(x) is negative for $\{x:-3<x<3\}$

Explanation of Solution

Given Information:

  $f(x)=x^2-9$ and f(x)< 0

Formula Used:

If

  $\begin{array}{l}{x}^2<a^2\\ Then\\ -a<x<a\end{array}$

Calculation:

Solving

  $\begin{array}{l}f(x)<0\\ f(x)=x^2-9<0\\ x^2-9+9<0+\mathrm{9...................}[add"9"each\_side]\\ x^2<9\\ -\sqrt{9}<\sqrt{x^2}<\sqrt{9}\mathrm{..................}[taking\_square\_root]\\ -3<x<3\end{array}$

x	$f(x)=x^2-9$
-4	7
-3	0
-2	-5
-1	-8
0	-9
1	-8
2	-5
3	0
4	7

So, f(x) is negative for $\{x:-3<x<3\}$

(d)

To determine

To Find:

The domain and range of $f(x)=\sqrt{x^2-9}$

Answer to Problem 66PPS

Domain = $\{x:x\le -3or3\le x\}$

Range = $\{y:y\ge 0\}$

Explanation of Solution

Given Information:

  $f(x)=\sqrt{x^2-9}$

Formula Used:

If

$\begin{array}{l}{x}^2\ge a^2\\ Then\\ x\le -a\\ or\\ x\ge a\end{array}$

$\begin{array}{l}If\\ y=\sqrt{x}\mathrm{...............}where\_x\_is\_real\_number\\ Then,\\ x\ge 0\\ and\\ \sqrt{x}\ge \mathrm{0................}where\_x\_is\_real\_number\end{array}$

Calculation:

Since $f(x)=\sqrt{x^2-9}$ is a positive square root .

  $\begin{array}{l}f(x)=\sqrt{x^2-9}\\ \Rightarrow x^2-9\ge 0\\ \Rightarrow x^2-9+9\ge 0+\mathrm{9..................}[add"9"each\_side]\\ \Rightarrow x^2\ge 9\\ \Rightarrow x^2\ge 3^2\\ \Rightarrow x\le -3\\ or\\ x\ge 3\end{array}$

So, the domain = $\{x:x\le -3or3\le x\}$

Since $f(x)=\sqrt{x^2-9}$ is a positive square root.

So , $f(x)=\sqrt{x^2-9}\ge 0$

Hence , the range is all numbers greater than or equal to 0.

Range = $\{y:y\ge 0\}$