GMAT Math : Understanding functions

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Example Questions

Example Question #51 : Functions/Series

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 even because it is a polynomial of degree 3.

is odd because it is a polynomial of degree 3.

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

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.

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Example Question #52 : Functions/Series

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

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

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

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Example Question #53 : Functions/Series

$\left \lfloor N \right \rfloor$ 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:

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

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

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

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

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

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

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Example Question #1278 : Gmat Quantitative Reasoning

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

Possible Answers:

$10-3\sqrt{2}$

$9-\sqrt{3}$

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):

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

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Example Question #55 : Functions/Series

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:

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

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

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

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

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

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 .

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Example Question #54 : Functions/Series

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:

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

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

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

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

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

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

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

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Example Question #53 : 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:

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

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

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

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

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

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

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Example Question #58 : Functions/Series

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:

$(-\infty, \infty)$

$[-27, \infty)$

$[6, \infty)$

$[-3, \infty)$

$[ 3, \infty)$

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 $[-3, \infty)$

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Example Question #54 : 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:

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

$\frac{1}{8}$

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

None of the other answers.

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

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

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

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Example Question #55 : 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:

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

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

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

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

The correct answer is not given among the other responses.

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))

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

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

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

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

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

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GMAT Math : Understanding functions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #51 : Functions/Series

Example Question #52 : Functions/Series

Example Question #53 : Functions/Series

Example Question #1278 : Gmat Quantitative Reasoning

Example Question #55 : Functions/Series

Example Question #54 : Functions/Series

Example Question #53 : Understanding Functions

Example Question #58 : Functions/Series

Example Question #54 : Understanding Functions

Example Question #55 : Understanding Functions

All GMAT Math Resources

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