We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

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

Study concepts, example questions & explanations for SAT Math

Create An Account

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Which of the following values of x is not in the domain of the function y = (2x – 1) / (x² – 6x + 9) ?

Possible Answers:

1/2

–1/2

Correct answer:

Explanation:

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.

Report an Error

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Given the relation below:

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

Find the range of the inverse of the relation.

Possible Answers:

{2, 4, 6, 8}

{1, 2, 3, 4}

{1, 3, 5, 7}

{5, 6, 7, 8}

the inverse of the relation does not exist

Correct answer: {1, 3, 5, 7}

Explanation:

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.

Report an Error

Example Question #4 : How To Find Domain And Range Of The Inverse Of A Relation

What is the range of the function y = x² + 2?

Possible Answers:

{2}

all real numbers

y ≥ 2

{–2, 2}

undefined

Correct answer:

y ≥ 2

Explanation:

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 = x² + 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 = x² + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.

Report an Error

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

$\dpi{100} -2$

$\dpi{100} 4$

$\dpi{100} 0$

$\dpi{100} 2$

$\dpi{100} -4$

Correct answer:

$\dpi{100} -2$

Explanation:

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 $\dpi{100} -2$ .

Report an Error

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

$f^{-1}(x)=\frac{1}{x-3}$

$f^{-1}(x)=x+3$

$f^{-1}(x)=3x$

$f^{-1}(x)=x-3$

$f^{-1}(x)=3-x$

Correct answer:

$f^{-1}(x)=x+3$

Explanation:

$f(x) = x - 3$ 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 $y=x+3$ .

Report an Error

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

$\frac{3x}{x-1}$

$\frac{-x}{x-3}$

$\frac{-x}{x+3}$

$\frac{-3x}{x+1}$

Correct answer:

$\frac{3x}{x-1}$

Explanation:

Inversef2

Inverse3

Report an Error

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

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

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

Possible Answers:

The inverse of the relation does not exist.

$\left \{ 5,\ 6,\ 7,\ 8 \right \}$

$\left \{ 1,\ 2,\ 3,\ 4 \right \}$

$\left \{ 1,\ 3,\ 5,\ 7 \right \}$

$\left \{ 2,\ 4,\ 6,\ 8 \right \}$

Correct answer:

$\left \{ 2,\ 4,\ 6,\ 8 \right \}$

Explanation:

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

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 $\left \{ 2,\ 4,\ 6,\ 8 \right \}$ .

Report an Error

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

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).

Possible Answers:

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

$f^{-1} (x)$ does not exist

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

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

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

Correct answer:

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

Explanation:

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 f(a) = f(b) , then a = b - or, equivalently, if there does not exist and such that $a \ne b$ , 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 $a \ne b$ such that f(a) = f(b) 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 a = 1, b = -4 :

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

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

To determine the inverse of f(x) , first, rewrite f(x) in vertex form

f(x) = a (x-h)+k

a = 1 , 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, h = 2, k=8 .

The vertex form of f(x) is therefore

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

To find $f^{-1}(x)$ , first replace f(x) 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 f(x) 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 $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$ .

Report an Error

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

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?

Possible Answers:

$x \in [3,9]$

$x \in [-15, -9]$

$x \in [9, 15 ]$

$x \in [-3, 3]$

$x \in [-9,- 3]$

Correct answer:

$x \in [-9,- 3]$

Explanation:

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.

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 a = 1, b = 9 :

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

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.

Report an Error

Example Question #41 : Algebraic Functions

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).

Possible Answers:

$f^{-1} (x)$ does not exist

$f^{-1} (x) = 8- \sqrt{2x+98}$

$f^{-1} (x) = 8- \sqrt{2x-98}$

$f^{-1} (x) = 8+ \sqrt{2x-98}$

$f^{-1} (x) = 8+ \sqrt{2x+98}$

Correct answer:

$f^{-1} (x)$ does not exist

Explanation:

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 $a \ne b$ , but f(a) = f(b) . 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 f(x) changes direction on the interval, and $f^{-1} (x)$ does not exist on $\left [ 0, \infty \right )$ .

Report an Error

View SAT Mathematics Tutors

Noah
Certified Tutor

Bard College, Bachelors, Philosophy and Creative Writing (Dual Major). Sarah Lawrence College, Master of Fine Arts, Creative ...

View SAT Mathematics Tutors

Priyanka
Certified Tutor

CUNY Hunter College, Bachelor in Arts, Economics.

View SAT Mathematics Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

SAT Math Tutors in Top Cities:

Atlanta SAT Math Tutors, Austin SAT Math Tutors, Boston SAT Math Tutors, Chicago SAT Math Tutors, Dallas Fort Worth SAT Math Tutors, Denver SAT Math Tutors, Houston SAT Math Tutors, Kansas City SAT Math Tutors, Los Angeles SAT Math Tutors, Miami SAT Math Tutors, New York City SAT Math Tutors, Philadelphia SAT Math Tutors, Phoenix SAT Math Tutors, San Diego SAT Math Tutors, San Francisco-Bay Area SAT Math Tutors, Seattle SAT Math Tutors, St. Louis SAT Math Tutors, Tucson SAT Math Tutors, Washington DC SAT Math Tutors

Popular Courses & Classes

Spanish Courses & Classes in Boston, GRE Courses & Classes in Chicago, SAT Courses & Classes in New York City, SSAT Courses & Classes in Boston, MCAT Courses & Classes in Denver, ACT Courses & Classes in Philadelphia, SSAT Courses & Classes in Houston, ISEE Courses & Classes in Atlanta, MCAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Philadelphia, ISEE Test Prep in San Diego, SAT Test Prep in Boston, SAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Washington DC, SSAT Test Prep in Seattle, MCAT Test Prep in Boston, SSAT Test Prep in Washington DC, SSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

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

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #4 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #41 : Algebraic Functions

All SAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: