How to find domain and range of the inverse of a relation - SAT Math

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Question

Which of the following values of x is not in the domain of the function y = (2_x –_ 1) / (x_2 – 6_x + 9) ?

Answer

Values of x that make the denominator equal zero are not included in the domain. The denominator can be simplified to (x – 3)2, so the value that makes it zero is 3.

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Question

Given the relation below, identify the domain of the inverse of the relation.

$\left {(1, 2), (3, 4), (5, 6), (7, 8){ \right }$

Answer

$\left {(1, 2), (3, 4), (5, 6), (7, 8){ \right }$

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

For the original relation, the range is: $\left { 2,\ 4,\ 6,\ 8 \right }$ .

Thus, the domain for the inverse relation will also be $\left { 2,\ 4,\ 6,\ 8 \right }$ .

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Question

Given the relation below:

{(1, 2), (3, 4), (5, 6), (7, 8)}

Find the range of the inverse of the relation.

Answer

The domain of a relation is the same as the range of the inverse of the relation. In other words, the x-values of the relation are the y-values of the inverse.

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Question

If $f(x)=\frac{x}{x-3}$ , then which of the following is equal to $f^{-1}(x)$ ?

Answer

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Question

What is the range of the function y = _x_2 + 2?

Answer

The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)

So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.

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Question

What is the smallest value that belongs to the range of the function $f(x)=2|4-x|-2$ ?

Answer

We need to be careful here not to confuse the domain and range of a function. The problem specifically concerns the range of the function, which is the set of possible numbers of $\dpi{100} f(x)$ . It can be helpful to think of the range as all the possible y-values we could have on the points on the graph of $\dpi{100} f(x)$ .

Notice that $\dpi{100} f(x)$ has $\dpi{100} |4-x|$ in its equation. Whenever we have an absolute value of some quantity, the result will always be equal to or greater than zero. In other words, |4-x| $\geq$ 0. We are asked to find the smallest value in the range of $\dpi{100} f(x)$ , so let's consider the smallest value of $\dpi{100} |4-x|$ , which would have to be zero. Let's see what would happen to $\dpi{100} f(x)$ if $\dpi{100} |4-x|=0$ .

$\dpi{100} f(x)=2(0)-2=0-2=-2$

This means that when $\dpi{100} |4-x|=0$ , $\dpi{100} f(x)=-2$ . Let's see what happens when $\dpi{100} |4-x|$ gets larger. For example, let's let $\dpi{100} |4-x|=3$ .

$\dpi{100} f(x)=2(3)-2=4$

As we can see, as $\dpi{100} |4-x|$ gets larger, so does $\dpi{100} f(x)$ . We want $\dpi{100} f(x)$ to be as small as possible, so we are going to want $\dpi{100} |4-x|$ to be equal to zero. And, as we already determiend, $\dpi{100} f(x)$ equals $\dpi{100} -2$ when $\dpi{100} |4-x|=0$ .

The answer is $\dpi{100} -2$ .

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Question

If $f(x) = x - 3$ , then find $f^{-1}(x)$

Answer

$f(x) = x - 3$ is the same as $y= x - 3$ .

To find the inverse simply exchange $x$ and $y$ and solve for $y$ .

So we get $x=y-3$ which leads to $y=x+3$ .

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Question

Define $f(x) = x^{2}- 4x - 4$ , restricting the domain of the function to $\left ( -\infty , 0 \right ]$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Answer

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and only if, if , then - or, equivalently, if there does not exist and such that , but . This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be such that on this interval. The key is therefore to determine whether the interval to which the domain is restricted contains the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2}- 4x - 4$ can be found by setting :

$x = - \frac{-4}{2 \cdot 1} = \frac{4}{2} =2$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 2. However, $2 otin \left ( -\infty , 0 \right ]$ . Therefore, the domain at which is restricted does not include the vertex, and $f^{-1} (x)$ exists on this domain.

To determine the inverse of , first, rewrite in vertex form

, the same as in the standard form.

The graph of , if unrestricted, would have a vertex with -coordinate 2, and -coordinate

$f(2) = 2^{2}- 4 \cdot 2 - 4 = 4 - 8 - 4 =-8$ .

Therefore, .

The vertex form of is therefore

$f(x)= (x - 2)^{2}- 8$

To find $f^{-1}(x)$ , first replace with :

$y= (x - 2)^{2}- 8$

Switch and :

$x= (y - 2)^{2}- 8$

Solve for . First, add 8 to both sides:

$x+ 8 = (y - 2)^{2}- 8 + 8$

$x+ 8 = (y - 2)^{2}$

Take the square root of both sides:

$y - 2 = \pm \sqrt{ x+ 8 }$

Add 2 to both sides

$y - 2+ 2 = \pm \sqrt{ x+ 8 } + 2$

$y = 2 \pm \sqrt{ x+ 8 }$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= 2 \pm \sqrt{ x+ 8 }$

Either $f^{-1}(x)= 2 + \sqrt{ x+ 8 }$ or $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

The domain of is the set of nonpositive numbers; this is consequently the range of $f^{-1}(x)$ . $2 + \sqrt{ x+ 8 }$ can only have positive values, so the only possible choice for $f^{-1}(x)$ is $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$ .

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Question

Define a function $f(x) = x^{2} + 9 x -17$ .

It is desired that is domain be restricted so that have an inverse. Which of these domain restrictions would not achieve that goal?

Answer

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and only if, if , then - or, equivalently, if there does not exist and such that , but . This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be such that on this interval. The key is therefore to identify the interval that contains the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2} + 9 x -17$ can be found by setting :

$x = - \frac{9}{2 \cdot 1} = - \frac{9}{2} = -4 \frac{1}{2}$ .

Of the five intervals in the choices,

$-4 \frac{1}{2} \in [-9,- 3]$ ,

so $f ^{-1}$ cannot exist if is restricted to this interval. This is the correct choice.

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Question

Define $f(x) = \frac{1}{2}x^{2}- 8x - 17$ , restricting the domain of the function to $\left [ 0, \infty \right )$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Answer

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.

$f^{-1} (x)$ exists on a given domain interval if and only if there does not exist and on this domain such that , but . This will happen if the graph changes direction on the domain interval. The key is therefore to determine whether the given domain interval includes the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = \frac{1}{2}x^{2}- 8x - 17$ can be found by setting $a = \frac{1}{2},b= -8$ :

$x = - \frac{(-8)}{2\left ( \frac{1}{2} \right )} = -\frac{-8}{1} = 8$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 8. Since $8 \in \left [ 0, \infty \right )$ , the vertex is in the interior of the domain; as a consequence, the graph of changes direction on the interval, and $f^{-1} (x)$ does not exist on $\left [ 0, \infty \right )$ .

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Question

Which of the following values of x is not in the domain of the function y = (2_x –_ 1) / (x_2 – 6_x + 9) ?

Answer

Values of x that make the denominator equal zero are not included in the domain. The denominator can be simplified to (x – 3)2, so the value that makes it zero is 3.

Compare your answer with the correct one above

Question

Given the relation below, identify the domain of the inverse of the relation.

$\left {(1, 2), (3, 4), (5, 6), (7, 8){ \right }$

Answer

$\left {(1, 2), (3, 4), (5, 6), (7, 8){ \right }$

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

For the original relation, the range is: $\left { 2,\ 4,\ 6,\ 8 \right }$ .

Thus, the domain for the inverse relation will also be $\left { 2,\ 4,\ 6,\ 8 \right }$ .

Compare your answer with the correct one above

Question

Given the relation below:

{(1, 2), (3, 4), (5, 6), (7, 8)}

Find the range of the inverse of the relation.

Answer

The domain of a relation is the same as the range of the inverse of the relation. In other words, the x-values of the relation are the y-values of the inverse.

Compare your answer with the correct one above

Question

If $f(x)=\frac{x}{x-3}$ , then which of the following is equal to $f^{-1}(x)$ ?

Answer

Compare your answer with the correct one above

Question

What is the range of the function y = _x_2 + 2?

Answer

The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)

So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.

Compare your answer with the correct one above

Question

What is the smallest value that belongs to the range of the function $f(x)=2|4-x|-2$ ?

Answer

We need to be careful here not to confuse the domain and range of a function. The problem specifically concerns the range of the function, which is the set of possible numbers of $\dpi{100} f(x)$ . It can be helpful to think of the range as all the possible y-values we could have on the points on the graph of $\dpi{100} f(x)$ .

Notice that $\dpi{100} f(x)$ has $\dpi{100} |4-x|$ in its equation. Whenever we have an absolute value of some quantity, the result will always be equal to or greater than zero. In other words, |4-x| $\geq$ 0. We are asked to find the smallest value in the range of $\dpi{100} f(x)$ , so let's consider the smallest value of $\dpi{100} |4-x|$ , which would have to be zero. Let's see what would happen to $\dpi{100} f(x)$ if $\dpi{100} |4-x|=0$ .

$\dpi{100} f(x)=2(0)-2=0-2=-2$

This means that when $\dpi{100} |4-x|=0$ , $\dpi{100} f(x)=-2$ . Let's see what happens when $\dpi{100} |4-x|$ gets larger. For example, let's let $\dpi{100} |4-x|=3$ .

$\dpi{100} f(x)=2(3)-2=4$

As we can see, as $\dpi{100} |4-x|$ gets larger, so does $\dpi{100} f(x)$ . We want $\dpi{100} f(x)$ to be as small as possible, so we are going to want $\dpi{100} |4-x|$ to be equal to zero. And, as we already determiend, $\dpi{100} f(x)$ equals $\dpi{100} -2$ when $\dpi{100} |4-x|=0$ .

The answer is $\dpi{100} -2$ .

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Question

If $f(x) = x - 3$ , then find $f^{-1}(x)$

Answer

$f(x) = x - 3$ is the same as $y= x - 3$ .

To find the inverse simply exchange $x$ and $y$ and solve for $y$ .

So we get $x=y-3$ which leads to $y=x+3$ .

Compare your answer with the correct one above

Question

Define $f(x) = x^{2}- 4x - 4$ , restricting the domain of the function to $\left ( -\infty , 0 \right ]$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Answer

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and only if, if , then - or, equivalently, if there does not exist and such that , but . This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be such that on this interval. The key is therefore to determine whether the interval to which the domain is restricted contains the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2}- 4x - 4$ can be found by setting :

$x = - \frac{-4}{2 \cdot 1} = \frac{4}{2} =2$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 2. However, $2 otin \left ( -\infty , 0 \right ]$ . Therefore, the domain at which is restricted does not include the vertex, and $f^{-1} (x)$ exists on this domain.

To determine the inverse of , first, rewrite in vertex form

, the same as in the standard form.

The graph of , if unrestricted, would have a vertex with -coordinate 2, and -coordinate

$f(2) = 2^{2}- 4 \cdot 2 - 4 = 4 - 8 - 4 =-8$ .

Therefore, .

The vertex form of is therefore

$f(x)= (x - 2)^{2}- 8$

To find $f^{-1}(x)$ , first replace with :

$y= (x - 2)^{2}- 8$

Switch and :

$x= (y - 2)^{2}- 8$

Solve for . First, add 8 to both sides:

$x+ 8 = (y - 2)^{2}- 8 + 8$

$x+ 8 = (y - 2)^{2}$

Take the square root of both sides:

$y - 2 = \pm \sqrt{ x+ 8 }$

Add 2 to both sides

$y - 2+ 2 = \pm \sqrt{ x+ 8 } + 2$

$y = 2 \pm \sqrt{ x+ 8 }$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= 2 \pm \sqrt{ x+ 8 }$

Either $f^{-1}(x)= 2 + \sqrt{ x+ 8 }$ or $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

The domain of is the set of nonpositive numbers; this is consequently the range of $f^{-1}(x)$ . $2 + \sqrt{ x+ 8 }$ can only have positive values, so the only possible choice for $f^{-1}(x)$ is $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$ .

Compare your answer with the correct one above

Question

Define a function $f(x) = x^{2} + 9 x -17$ .

It is desired that is domain be restricted so that have an inverse. Which of these domain restrictions would not achieve that goal?

Answer

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and only if, if , then - or, equivalently, if there does not exist and such that , but . This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be such that on this interval. The key is therefore to identify the interval that contains the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2} + 9 x -17$ can be found by setting :

$x = - \frac{9}{2 \cdot 1} = - \frac{9}{2} = -4 \frac{1}{2}$ .

Of the five intervals in the choices,

$-4 \frac{1}{2} \in [-9,- 3]$ ,

so $f ^{-1}$ cannot exist if is restricted to this interval. This is the correct choice.

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Question

Define $f(x) = \frac{1}{2}x^{2}- 8x - 17$ , restricting the domain of the function to $\left [ 0, \infty \right )$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Answer

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.

$f^{-1} (x)$ exists on a given domain interval if and only if there does not exist and on this domain such that , but . This will happen if the graph changes direction on the domain interval. The key is therefore to determine whether the given domain interval includes the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = \frac{1}{2}x^{2}- 8x - 17$ can be found by setting $a = \frac{1}{2},b= -8$ :

$x = - \frac{(-8)}{2\left ( \frac{1}{2} \right )} = -\frac{-8}{1} = 8$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 8. Since $8 \in \left [ 0, \infty \right )$ , the vertex is in the interior of the domain; as a consequence, the graph of changes direction on the interval, and $f^{-1} (x)$ does not exist on $\left [ 0, \infty \right )$ .

Compare your answer with the correct one above

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