Call Now to Set Up Tutoring:

(888) 888-0446

SAT Math : SAT Mathematics

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 #2871 : Sat Mathematics

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate (f-g)(19) .

Possible Answers:

$36 \frac{1}{2}$

$81\frac{1}{2}$

$5 \frac{1}{9}$

$-3\frac{1}{7}$

Correct answer:

$36 \frac{1}{2}$

Explanation:

By definition, (f-g)(19) = f(19)- g(19) ; simply evaluate f(19) and g(19) by substituting 19 for in both definitions, and subtract:

f(x) = 4x- 17

$f(19) = 4 \cdot 19 - 17 = 76 - 17 = 59$

$g(x) = \frac{1}{2}x+ 13$

$g(19) = \frac{1}{2} \cdot 19 + 13 = 9\frac{1}{2} + 13 = 22 \frac{1}{2}$

$(f-g)(19) = f(19)- g(19) = 59 - 22 \frac{1}{2} = 36 \frac{1}{2}$

Report an Error

Example Question #1094 : Algebra

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

$(f \circ g) (c) = 10$ .

Evaluate .

Possible Answers:

$2 \frac{3}{4}$

$-12 \frac{1}{2}$

$24 \frac{1}{2}$

None of the other choices gives the correct response.

Correct answer:

$-12 \frac{1}{2}$

Explanation:

By definition, $(f \circ g ) (x) = f [g(x)]$ .

Replacing g(x) with its definition, we get

$(f \circ g ) (x) = f \left ( \frac{1}{2}x+ 13 \right )$

In the definition of f ( x) , replace with $\frac{1}{2}x+ 13$ and simplify the expression:

f(x) = 4x- 17

$f \left ( \frac{1}{2}x+ 13 \right )= 4 \left ( \frac{1}{2}x+ 13 \right )- 17$

= 2 x+ 52 - 17

= 2 x+ 35

Therefore,

$(f \circ g ) (x) = 2 x+ 35$

$(f \circ g) (c) = 10$ ,

then

2 c+35 = 10

Solve for :

2 c+35-35 = 10 - 35

2 c = -25

$\frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot (-25)$

$c = -12 \frac{1}{2}$

Report an Error

Example Question #1095 : Algebra

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

$(g \circ f ) (c) = 10$ .

Evaluate .

Possible Answers:

$2 \frac{3}{4}$

$24 \frac{1}{2}$

None of the other choices gives the correct response.

$-12 \frac{1}{2}$

Correct answer:

$2 \frac{3}{4}$

Explanation:

By definition, $(g \circ f ) (x) = g [f(x)]$ .

Replacing f(x) with its definition, we get

$(g \circ f ) (x) = g ( 4x- 17)$

In the definition of g ( x) , replace with 4x-17 and simplify the expression:

$g(x) = \frac{1}{2}x+ 13$

$g(4x-17 ) = \frac{1}{2} (4x-17 )+ 13$

$= 2 x-8 \frac{1}{2} + 13$

$= 2 x+4 \frac{1}{2}$

Therefore,

$(g \circ f ) (x) = 2 x+4 \frac{1}{2}$

$(g \circ f ) (c) = 10$ ,

then

$2 c+4 \frac{1}{2} = 10$

Solve for :

$2 c+4 \frac{1}{2} - 4 \frac{1}{2} = 10 - 4 \frac{1}{2}$

$2 c = 5 \frac{1}{2}$

$\frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot 5 \frac{1}{2}$

$c = 2 \frac{3}{4}$

Report an Error

Example Question #161 : Algebraic Functions

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

(f-g) (c) = 19

Evaluate .

Possible Answers:

c =14

$c = 81\frac{1}{2}$

$c = 5 \frac{1}{9}$

$c = 36 \frac{1}{2}$

$c = -3\frac{1}{7}$

Correct answer:

c =14

Explanation:

To obtain the definition of the function (f-g)(x) , subtract the expressions that define the individual functions f(x) and g(x) :

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

$(f+g)(x)= (4x- 17 )- \left ( \frac{1}{2}x+ 13 \right )$

$= 4x- \frac{1}{2}x- 17 -13$

$= \frac{7}{2}x- 30$

(f+g) (c) = 19 , so

$\frac{7}{2}c- 30 = 19$

Solve for ; add 30:

$\frac{7}{2}c- 30 + 30 = 19 + 30$

$\frac{7}{2}c = 49$

Multiply by $\frac{2}{7}$ :

$\frac{2} {7}\cdot \frac{7}{2}c = \frac{2} {7}\cdot 49$

c =14

Report an Error

Example Question #166 : Algebraic Functions

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

(f+g) (c) = 19

Evaluate .

Possible Answers:

$c = 36 \frac{1}{2}$

$c = 5 \frac{1}{9}$

$c = 81\frac{1}{2}$

c =14

$c = -3\frac{1}{7}$

Correct answer:

$c = 5 \frac{1}{9}$

Explanation:

To obtain the definition of the function (f+g)(x) , add the expressions that define the individual functions f(x) and g(x) "

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

$(f+g)(x)= (4x- 17 )+ \left ( \frac{1}{2}x+ 13 \right )$

$= 4x+ \frac{1}{2}x- 17 + 13$

$= \frac{9}{2}x- 4$

(f+g) (c) = 19 , so

$\frac{9}{2}c- 4 = 19$ ;

Solve for ; add 4;

$\frac{9}{2}c- 4+ 4 = 19 + 4$

$\frac{9}{2}c = 23$

Multiply by $\frac{2}{9}$ :

$\frac{2}{9} \cdot \frac{9}{2}c =\frac{2}{9} \cdot 23$

$c =\frac{46}{9} = 5 \frac{1}{9}$

Report an Error

Example Question #1095 : Algebra

Define f(x) = 7x - 17 , restricting the domain to $x \in [-4, 8]$ .

Give the range of .

Possible Answers:

$\left [ -45, 39 \right ]$

$\left [ -147, -63\right ]$

$\left [ -45, \infty \right )$

$\left [ -147, \infty \right )$

$(-\infty, \infty)$

Correct answer:

$\left [ -45, 39 \right ]$

Explanation:

A function of the form f(x) = m x + b is a linear function and is either constantly increasing or constantly decreasing. Therefore, f(x) = 7x - 17 has its minimum and maximum values at the endpoints of its domain.

We evaluate f(-4) and f(8) by substitution, as follows:

f(x) = 7x - 17

f(-4) = 7 (-4) - 17 = -28 - 17 = -45

f(8) = 7 (8) - 17 = 56 - 17 = 39

The range of the function on the domain to which it is restricted is $\left [ -45, 39 \right ]$ .

Report an Error

Example Question #171 : Algebraic Functions

Define the function f(x) as follows:

$f(x) = \left\{\begin{matrix} 2x- 7& \textrm{ for } x < 0 \\ 4x - 14 & \textrm{ for } x \ge 0 \end{matrix}\right.$

Give the range of f(x) .

Possible Answers:

$(-\infty ,-14] \cup (-7, \infty)$

$(-\infty , -7)$

$[-14, \infty)$

[-14, -7)

$(-\infty , \infty)$

Correct answer:

$(-\infty , \infty)$

Explanation:

Since the piecewise-defined function f(x) is defined two different ways, one for negative numbers and one for nonnegative numbers, examine both definitions and determine each partial range separately; the union of the partial ranges will be the overall range.

If x < 0 , then

f(x ) = 2x- 7

Since

x < 0 ,

applying the properties of inequality,

$2 x < 2 \cdot 0$

2 x < 0

2 x - 7 < 0 - 7

2 x - 7 < - 7

f(x) < -7

Therefore, on the portion of the domain comprising nonpositive numbers, the partial range of f(x) is the set $(-\infty , -7)$ .

If $x \ge 0$ , then

f(x ) = 4x-14

Since

$x \ge 0$ ,

applying the properties of inequality,

$4 x \ge 4 \cdot 0$

$4 x \ge 0$

$4 x -14 \ge 0 - 14$

$4 x -14 \ge - 14$

$f(x) \ge - 14$

Therefore, on the portion of the domain comprising positive numbers, the partial range of f(x) is the set $[-14, \infty)$ .

The overall range is the union of these partial ranges, which is $(-\infty , \infty)$ .

Report an Error

Example Question #171 : Algebraic Functions

Define $f(x) = x^{2}- 8x - 4$ , restricting the domain of the function to $\left [ 0, \infty \right )$ .

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

Possible Answers:

$f^{-1}(x)= 4+ \sqrt{ x-20 }$

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

$f^{-1}(x)= 4+ \sqrt{ x+ 20 }$

$f^{-1}(x)= 4 - \sqrt{ x+ 20 }$

$f^{-1}(x)= 4 - \sqrt{ x- 20 }$

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 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 $f(x) = x^{2}- 8x - 4$ can be found by setting a = 1, b = -8 :

$x = - \frac{-8}{2 \cdot 1} = \frac{8}{2} =4$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 4. Since $4 \in \left [ 0, \infty \right )$ , the vertex is in the interior of the domain; as a consequence, $f^{-1} (x)$ does not exist on $\left [ 0, \infty \right )$ .

Report an Error

Example Question #1103 : Algebra

Define f(x) = -4x + 12 , restricting the domain to $x \in [5 , \infty )$ .

Give the range of f(x) .

Possible Answers:

$[- 8, \infty)$

$\left ( -\infty, -68 \right ]$

$\left ( -\infty,\infty)$

$[- 6 8, \infty)$

$\left ( -\infty, -8 \right ]$

Correct answer:

$\left ( -\infty, -8 \right ]$

Explanation:

A function of the form f(x) = m x + b is a linear function and is either constantly increasing or constantly decreasing. Therefore, we can simply note that if

$x \ge 5$ ,

as stated in the domain, then, multiplying both sides by , remembering to switch the symbol since we are multiplying by a negative number:

$-4 \cdot x \le -4 \cdot 5$

$-4 x \le -20$

Add 12 to both sides:

$-4 x + 12 \le -20 + 12$

$-4 x + 12 \le -8$

Replacing, we see that

$f(x) \le -8$ ,

so the range of f(x) is $\left ( -\infty, 8 \right ]$ .

Report an Error

Example Question #172 : Algebraic Functions

Define $f(x) = \frac{1}{2x+3}$ , restricting the domain of the function to the interval $[1, \infty)$ .

Give the range of the function.

Possible Answers:

$\left (0, \frac{1}{5} \right ]$

$\left (0, \infty \right )$

$\left (- \infty , \infty \right )$

$\left (-\infty , \frac{1}{5} \right ]$

$\left [\frac{1}{5} , \infty \right )$

Correct answer:

$\left (0, \frac{1}{5} \right ]$

Explanation:

If $x \ge 1$ , it follows by applying the properties of inequality that:

$2 \cdot x \ge 2 \cdot 1$

$2x \ge 2$

$2x + 3 \ge 2 + 3$

$2x + 3 \ge 5$

Multiply both sides by $\frac{1}{5(2x+3)}$ , which must be positive by closure:

$(2x + 3 )\cdot \frac{1}{5(2x+3)} \ge 5 \cdot \frac{1}{5(2x+3)}$

$\frac{1}{5 } \ge \frac{1}{2x+3}$ ,

that is,

$\frac{1}{2x+3} \le \frac{1}{5 }$

Also,by closure, $\frac{1}{2x+3} > 0$ , so

$0 < \frac{1}{2x+3} \le \frac{1}{5 }$

$0 < f(x) \le \frac{1}{5 }$

This makes the correct range $\left (0, \frac{1}{5} \right ]$ .

Report an Error

View SAT Mathematics Tutors

Zachary
Certified Tutor

Yale University, Bachelor of Science, Biology, General.

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

Cole
Certified Tutor

UW Madison, Bachelors, Industrial Engineering. UW Madison, Masters, Industrial Engineering.

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

GRE Courses & Classes in Seattle, SSAT Courses & Classes in Boston, Spanish Courses & Classes in Chicago, SAT Courses & Classes in Chicago, LSAT Courses & Classes in Los Angeles, SAT Courses & Classes in Denver, ACT Courses & Classes in Washington DC, SSAT Courses & Classes in Chicago, GRE Courses & Classes in Chicago, GMAT Courses & Classes in Atlanta

Popular Test Prep

GRE Test Prep in San Diego, GRE Test Prep in Phoenix, ACT Test Prep in Washington DC, LSAT Test Prep in Atlanta, ISEE Test Prep in Philadelphia, SAT Test Prep in Denver, SAT Test Prep in Los Angeles, LSAT Test Prep in New York City, ACT Test Prep in Philadelphia, LSAT Test Prep in San Diego

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 : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2871 : Sat Mathematics

Example Question #1094 : Algebra

Example Question #1095 : Algebra

Example Question #161 : Algebraic Functions

Example Question #166 : Algebraic Functions

Example Question #1095 : Algebra

Example Question #171 : Algebraic Functions

Example Question #171 : Algebraic Functions

Example Question #1103 : Algebra

Example Question #172 : Algebraic Functions

All SAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: