GMAT Math : Understanding functions

Study concepts, example questions & explanations for GMAT Math

Create An Account

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1275 : Problem Solving Questions

Consider the function $f(x) = x^{3} - 2$ .

State whether this function is even, odd, or neither, and give the reason for your answer.

Possible Answers:

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

is odd because f(-x) = -f(x) for each value of in the domain.

is odd because it is a polynomial of degree 3.

is even because f(-x) = f(x) for each value of in the domain.

is even because it is a polynomial of degree 3.

Correct answer:

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

Explanation:

A function is odd if and only if f(-x) = -f(x) for each value of in the domain; it is even if and only if f(-x) = f(x) for each value of in the domain. To disprove a function is odd or even, we need only find one value of for which the appropriate statement fails to hold.

Consider x = 1 :

$f(x) = x^{3} - 2$

$f(1) = 1^{3} - 2 = 1-2 = -1$

$f(x) = x^{3} - 2$

$f(-1) = (-1)^{3} - 2 = -1 -2 = -3$

$f(-1 ) \neq - f(1)$ , so is not an odd function; $f(-1 ) \neq f(1)$ , so is not an even function.

Report an Error

Example Question #1276 : Problem Solving Questions

$f (x) = \left\{\begin{matrix} 1\textrm{ if } x< 0\\ x \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} 2x-5 \textrm{ if } x< 0\\ 2\; \; \; \; \; \; \; \; \textrm{ if } x \geq 0 \end{matrix}\right.$ .

Evaluate $(f \circ g) (-4)$ .

Possible Answers:

Correct answer:

Explanation:

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

First we evaluate g (-4) . Since the parameter is negative, we use the first half of the definition of :

g (x) = 2x-5

$g (-4) = 2 \left ( -4\right )-5 = -8 - 5 = -13$

f (g(-4)) = f(-13) ; since the parameter here is again negative, we use the first half of the definition of :

f(-13) = 1

Therefore, $(f \circ g) (-4) = 1$ .

Report an Error

Example Question #1277 : Problem Solving Questions

$\left \lfloor N \right \rfloor$ is defined to be the greatest integer less than or equal to .

Define $f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$ .

Evaluate $f \left ( \frac{3}{5} \right )$ .

Possible Answers:

$f \left ( \frac{3}{5} \right ) = 8$

$f \left ( \frac{3}{5} \right ) = 6$

$f \left ( \frac{3}{5} \right ) = 5$

$f \left ( \frac{3}{5} \right ) = 9$

$f \left ( \frac{3}{5} \right ) = 7$

Correct answer:

$f \left ( \frac{3}{5} \right ) = 6$

Explanation:

$f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$

$f \left ( \frac{3}{5} \right ) = \left \lfloor \left (\frac{3}{5} \right ) ^{2} \right \rfloor - \left \lfloor 4 \cdot \frac{3}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor \frac{12}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor 2 \frac{2}{5}\right \rfloor + 8$

=0 - 2 + 8 = 6

Report an Error

Example Question #54 : Understanding Functions

If $f(x)=x^{2}-\sqrt{x}$ and g(x)=3x-2 , what is f(g(2)) ?

Possible Answers:

$9-\sqrt{3}$

$10-3\sqrt{2}$

Correct answer:

Explanation:

We start by finding g(2):

$g(2)=3\cdot2-2 = 6-2 = 4$

Then we find f(g(2)) which is f(4):

$f(4)=4^{2}-\sqrt{4}=16-2=14$

Report an Error

Example Question #55 : Understanding Functions

Define two real-valued functions as follows:

$p (x)= x^{2}$

$q(x )= \left\{\begin{matrix} x+1 & x< 0\\ x-1 & x \ge 0 \end{matrix}\right.$

Determine $\left ( q \circ p\right )(x)$ .

Possible Answers:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+2x + 1 & x < 0\\ x^{2}-2x + 1&x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x)= x^{2}-2x + 1$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+ 1 & x < 0\\ x^{2}- 1&x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x)= x^{2}+1$

$\left ( q \circ p\right )(x)= x^{2}- 1$

Correct answer:

$\left ( q \circ p\right )(x)= x^{2}- 1$

Explanation:

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2})$ by definition. is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, $p(x)= x^{2}$ is nonnegative for all real numbers, so the defintion for nonnegative numbers, q(x )= x-1 , is the one that will always be used. Therefore,

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2}) = x^{2}-1$ for all values of .

Report an Error

Example Question #56 : Understanding Functions

Define two real-valued functions as follows:

$p (x)= 9 - x^{2}$

$q(x )= \left\{\begin{matrix} 0 & x< 0\\ x & x \ge 0 \end{matrix}\right.$

Determine $\left ( q \circ p\right )(x)$ .

Possible Answers:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

$\left ( q \circ p\right )(x) = 9 - x^{2}$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 9 & x< 0\\ 9-x^{2} &x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 9 - x^{2}& x < -3 \\ 0& -3 \le x \le 3 \\ 9 - x^{2}& x > 3 \end{matrix}\right.$

$\left ( q \circ p\right )(x) =0$

Correct answer:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

Explanation:

$\left ( q \circ p\right )(x) = q(p(x))= q( 9 - x^{2})$ by definition.

is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values.

If $9 -x^{2} < 0$ , then we use the definition q(x)= 0 . This happens if

$9 -x^{2} < 0$

$9 < x^{2}$

x > 3 or x < -3

Therefore, the defintion of $\left ( q \circ p\right )(x)$ for x > 3 or x < -3 is

$\left ( q \circ p\right )(x) = 0$

Subsquently, if $-3 \le x \le 3$ , we use the defintion q(x) = x , since $9 - x^{2} \ge 0$ :

$\left ( q \circ p\right )(x) =q( 9 - x^{2}) = 9 -x^{2}$ .

The correct choice is

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

Report an Error

Example Question #57 : Understanding Functions

Define a function on the real numbers as follows:

$f(x) = x +4\sqrt{x}- 6$

Give the range of the function.

Possible Answers:

$\left [ 2, \infty \right )$

$\left [ -6, \infty \right )$

$\left [ 4, \infty \right )$

$\left [ -10, \infty \right )$

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

Correct answer:

$\left [ -6, \infty \right )$

Explanation:

This can be understood better by substituting $t = \sqrt{x}$ , and, subsequently, $t^{2} =\left ( \sqrt{x} \right )^{2} = x$ in the function's definition.

$x + 4 \sqrt{x}- 6 = t^{2} +4 t - 6$

which is now in standard quadratic form in terms of .

Write this in vertex form by completing the square:

$t^{2} +4 t - 6 = t^{2}+4 t + \left ( \frac{4}{2} \right ) ^{2} - 6 - \left ( \frac{4}{2} \right ) ^{2}$

$= t^{2}+4 t +4- 6 - 4$

$=\left ( t+2 \right )^{2}- 10$

Substitute $\sqrt{x}$ back for , and the original function can be rewritten as

$f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10$ .

To find the range, note that $\sqrt{x} \ge 0$ . Therefore,

$\sqrt{x} + 2 \ge 2$

$\left ( \sqrt{x}+2 \right )^{2} \ge 4$

and

$f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10 \ge -6$

The range of is the set $\left [ -6, \infty \right )$ .

Report an Error

Example Question #58 : Understanding Functions

Define a function on the real numbers as follows:

$f(x) = x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6$

Give the range of the function.

Possible Answers:

$[ 3, \infty)$

$(-\infty, \infty)$

$[6, \infty)$

$[-27, \infty)$

$[-3, \infty)$

Correct answer:

$[-3, \infty)$

Explanation:

This can be understood better by substituting $t = x^{\frac{1}{3}}$ , and, subsequently, $t^{2} =\left ( x^{\frac{1}{3}} \right )^{2} = x^{\frac{2}{3}}$ in the function's definition.

$x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6 = t^{2}+ 6t + 6$

which is now in standard quadratic form in terms of .

Write this in vertex form by completing the square:

$t^{2}+ 6t + 6$

$= t^{2}+ 6t +\left ( \frac{6}{2} \right )^{2}+ 6 - \left ( \frac{6}{2} \right )^{2}$

$= t^{2}+ 6t +9+ 6 - 9$

$= (t+3)^{2} - 3$

Substitute $x^{\frac{1}{3}}$ back for . The original function can be rewritten as

$f(x)= (x^{\frac{1}{3}}+3)^{2} - 3$

or, in radical form,

$f(x)= (\sqrt[3]{x }+3)^{2} - 3$

$\sqrt[3]{x}$ can assume any real value; so, subsequently, can $\sqrt[3]{x} + 3$ . But its square must be nonnegative, so

$\left (\sqrt[3]{x} + 3 \right )^{2} \ge 0$

and

$f(x)= (\sqrt[3]{x }+3)^{2} - 3 \ge -3$

The range of is $[-3, \infty)$

Report an Error

Example Question #59 : Understanding Functions

If the functions f(x)=2x^2+a and g(x) =-x^2-b intersect only when x=1 , and that f(g(0)) =1 , and b>0 , what is ?

Possible Answers:

$\frac{1}{4}(1+\sqrt{33})$

$\frac{1}{8}$

$\frac{1}{8}(1-\sqrt{33})$

$\frac{1}{8}(3+\sqrt{99})$

None of the other answers.

Correct answer:

$\frac{1}{4}(1+\sqrt{33})$

Explanation:

In order to find exactly the a,b values where the equations intersect and when f(g(0))=1 . We need to consider each piece of information seperately.

Let's start with f(g(0))=1 . Plugging into , we have f(g(x)) = 2(-x^2-b)^2+a . Plugging 0 into this, we have f(g(0))=2b^2+a . This in turn equals 1, because we were given that piece of information in the beginning. So we end up with 2b^2+a=1

Now let's shift our attention to "intersect only when x=1 " That means, if we plug 1 into both equations, we can set them equal to each other.

2(1)^2+a=-(1)^2-b becomes 2+a = -1-b becomes a+b= -3 .

Now we have two different equations arising from the two previous paragraphs.

a+b = -3

a+2b^2=1

We can solve this system of equations using the substitution method.

Solving for in the first equation gives a=-3-b .

Plugging this equation in for the 2nd equation gives (-3-b) +2b^2 =1 . Using algebra on this equation we get 2b^2-b-4=0

Now we are ready to use the quadratic formula to solve for .

$b = \frac{1\pm \sqrt{1^2-(4)(2)(-4)}}{2(2)}=\frac{1}{4}(1\pm \sqrt{33})$

Finally, since we're told in the beginning that b>0 , we must pick the plus sign in our solution for . Hence

$b =\frac{1}{4}(1+ \sqrt{33})$ .

Report an Error

Example Question #60 : Understanding Functions

Define two real-valued functions as follows:

$f(x) = x^{2} - 6$

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

Determine $\left ( f \circ g\right )(x)$ .

Possible Answers:

The correct answer is not given among the other responses.

$\left ( f \circ g\right )(x)= \frac{1 - 24x^{2}}{4x^{2}}$

$\left ( f \circ g\right )(x)= 2x^{3}- 12x$

$\left ( f \circ g\right )(x)= \frac{x^{2} - 6}{2x}$

$\left ( f \circ g\right )(x) = \frac{1}{2x^{2}-12}$

Correct answer:

$\left ( f \circ g\right )(x)= \frac{1 - 24x^{2}}{4x^{2}}$

Explanation:

This question is asking us to find the composition of f and g. In order to do this we need to plug g(x) into the x value in f(x).

$\left ( f \circ g\right )(x)$

= f(g(x))

$= f \left ( \frac{1}{2x} \right )$

$= \left ( \frac{1}{2x} \right )^{2} - 6$

$= \frac{1}{4x^{2}} - 6$

$= \frac{1}{4x^{2}} - \frac{24x^{2}}{4x^{2}}$

$= \frac{1 - 24x^{2}}{4x^{2}}$

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ACT Tutors in San Francisco-Bay Area, SSAT Tutors in Washington DC, Reading Tutors in Seattle, Algebra Tutors in Washington DC, Biology Tutors in Chicago, Reading Tutors in Philadelphia, English Tutors in Philadelphia, Biology Tutors in Houston, Biology Tutors in Philadelphia, ISEE Tutors in Atlanta

Popular Courses & Classes

ISEE Courses & Classes in Miami, GMAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Miami, SAT Courses & Classes in Atlanta, ISEE Courses & Classes in Houston, SSAT Courses & Classes in Boston, GRE Courses & Classes in Philadelphia, MCAT Courses & Classes in Phoenix, GRE Courses & Classes in New York City, ACT Courses & Classes in Philadelphia

Popular Test Prep

LSAT Test Prep in Washington DC, MCAT Test Prep in Philadelphia, GRE Test Prep in Miami, MCAT Test Prep in Phoenix, SAT Test Prep in Miami, GMAT Test Prep in Philadelphia, GMAT Test Prep in New York City, ISEE Test Prep in Seattle, LSAT Test Prep in New York City, ACT Test Prep in Los Angeles

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)

Email address:
Your name:
Feedback:

GMAT Math : Understanding functions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1275 : Problem Solving Questions

Example Question #1276 : Problem Solving Questions

Example Question #1277 : Problem Solving Questions

Example Question #54 : Understanding Functions

Example Question #55 : Understanding Functions

Example Question #56 : Understanding Functions

Example Question #57 : Understanding Functions

Example Question #58 : Understanding Functions

Example Question #59 : Understanding Functions

Example Question #60 : Understanding Functions

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details