Chapter 9.1, Problem 66PPS

a.

To determine

To state a domain for the given equation.

Answer to Problem 66PPS

As x can take up any real values, the domain is $\left\{-\infty <x<\infty \right\}$ ..

Explanation of Solution

Given information :

The function provided is $f(x)=x^2-9$

Since the given function is a quadratic one, the graph is that of a parabola. In case of a parabola, the domain is all values with x can take. Thus, the domain is $\left\{-\infty <x<\infty \right\}$

b.

To determine

To state a range for the given equation.

Answer to Problem 66PPS

The range is the values of y, it can only take up values greater than or equal to the minimum. That is, $\left\{y|y\ge -9\right\}$ .

Explanation of Solution

Given information :

The function provided is $f(x)=x^2-9$

Formula used :

Formula to compute the x-coordinate of the vertex is $x=-\frac{b}{2\times a}$ . Here, $a=1$ , $b=0$ and $c=-9$ . ‘a’ and ‘c’ are coefficients $x^2$ and $x^0$ respectively. The corresponding value of ‘y’ for the vertex would be the minimum.

Calculation :

Since the coefficient of ‘a’ is positive, the curve of this function is upward facing and hence the function has a minimum value.

X-coordinate of the vertex is

  $x=-\frac{b}{2\times a}$

Formula for axis of symmetry.

  $x=-\frac{0}{2\times 1}$

Putting the values of ‘a’ and ‘c’ .

  $x=0$

Vertex can be found out by putting the value of x computed in the axis of symmetry in the original function. This will give a value of $f(x)$ , that will be the maximum.

  $f(0)={(0)}^2-9$

Putting the value of $x=0$ in the original function.

Simplifying the expression.

  $f(0)=-9$

Thus, the minimum is at $-9$ .

Since x can take up any real values, the domain is $\left\{-\infty <x<\infty \right\}$ .

Since the range is the values of y, it can only take up values greater than or equal to the minimum. That is, $\left\{y|y\ge -9\right\}$ .

Since the given function is a quadratic one, the graph is that of a parabola. In case of a parabola, the domain is all values with x can take. Thus, the domain is $\left\{-\infty <x<\infty \right\}$ and the range is $\left\{y|y\ge -9\right\}$ .

c.

To determine

To calculate the values of x for which $f(x)$ is negative.

Answer to Problem 66PPS

The value of $f(x)$ is negative for $\{-3<x<3\}$ .

Explanation of Solution

Given information :

The function provided is $f(x)=x^2-9$

Formula used :

To calculate values of x for which $f(x)=0$ , use the formula to find roots of the quadratic equation. This can be found by using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . Here ‘a’, ‘b’ and ‘c’ are the co-efficient of the given equation.

Here, $a=1$ , $b=0$ and $c=-9$ .

Calculation :

  $x=\frac{-0\pm \sqrt{0^2-4\times (1)\times (-9)}}{2(1)}$

Putting the values in the formula.

  $x=\frac{-0\pm \sqrt{{36}^2}}{2}$

Simplifying

  $x=\frac{\pm 6}{2}$

Solving the two fractions.

  $x=3\text{ and }-3$

Thus, $f(x)=0$ between $(-3,0)\text{ and (3,0)}$ .

Since the given function is a quadratic one, the graph is that of a parabola. In case of a parabola, with the curve facing upwards all values of x between the two x intercepts will give a negative value for $f(x)$ .

d.

To determine

To state domain and range for the given equation.

Answer to Problem 66PPS

A domain for this function would be $\left\{x\le -3\right\}\text{ and }\left\{x\ge 3\right\}$ and the range would be $\left\{f(x)|f(x)\ge 0\right\}$ .

Explanation of Solution

Given information :

The function provided is $f(x)=\sqrt{x^2-9}$ .

Since the given function cannot be negative the range is $f(x)\ge 0$ . If $(x^2-9)$ is negative in the function $f(x)=\sqrt{x^2-9}$ , the value become undefined. Therefore, the domain is $\left\{x\le -3\right\}\text{ and }\left\{x\ge 3\right\}$ for which $f(x)$ is defined.