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College Algebra : Miscellaneous Functions

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Example Questions

Example Question #1 : Miscellaneous Functions

Define a function $f(x) = \sqrt{4x- 9}$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}$

$f^{-1}(x)= \frac{9}{4} x ^{2}+4$

$f^{-1}(x)= \frac{9}{4} x ^{2}+ \frac{1}{4}$

$f^{-1}(x)= \frac{1}{4} x ^{2}+9$

$f^{-1}(x)= \frac{9}{4} x ^{2}+9$

Correct answer:

$f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = \sqrt{4x- 9}$

$y= \sqrt{4x- 9}$

Switch the positions of and :

$x= \sqrt{4y- 9}$

$\sqrt{4y- 9} = x$

Solve for . This can be done as follows:

Square both sides:

$\left (\sqrt{4y- 9} \right ) ^{2}= x ^{2}$

$4y- 9 = x ^{2}$

Add 9 to both sides:

$4y- 9+9 = x ^{2}+9$

$4y = x ^{2}+9$

Multiply both sides by $\frac{1}{4}$ , distributing on the right:

$\frac{1}{4} \cdot 4y = \frac{1}{4} \cdot \left (x ^{2}+9 \right )$

$y = \frac{1}{4} \cdot x ^{2}+ \frac{1}{4} \cdot 9$

$y = \frac{1}{4} x ^{2}+ \frac{9}{4}$

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

$f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}$

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Example Question #2 : Miscellaneous Functions

Function

Refer to the above diagram, which shows the graph of a function f(x) .

True or false: $f\left ( \frac{1}{2} \right ) = 2$ .

Possible Answers:

False

True

Correct answer:

False

Explanation:

The statement is false. Look for the point on the graph of f(x) with -coordinate $\frac{1}{2}$ by going right $\frac{1}{2}$ unit, then moving up and noting the -value, as follows:

Function

$f \left ( \frac{1}{2} \right ) = \frac{3}{2}$ , so the statement is false.

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Example Question #1 : Miscellaneous Functions

Function

The above diagram shows the graph of function f(x) on the coordinate axes. True or false: The -intercept of the graph is (0, 3)

Possible Answers:

False

True

Correct answer:

False

Explanation:

The -intercept of the graph of a function is the point at which it intersects the -axis (the vertical axis). That point is marked on the diagram below:

Function

The point is about one and three-fourths units above the origin, making the coordinates of the -intercept $\left ( 0, 1\frac{3}{4}\right )$ .

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Example Question #171 : College Algebra

Function

A function f(x) is defined on the domain $\left \{ 1, 2, 3, 4, 5\right \}$ according to the above table.

Define a function g(x) = 3x- 5 . Which of the following values is not in the range of the function $(g \circ f )(x)$ ?

Possible Answers:

Correct answer:

Explanation:

This is the composition of two functions. By definition, $(g \circ f )(x) = g \left [ f(x)\right ]$ . To find the range of $(g \circ f )(x)$ , we need to find the values of this function for each value in the domain of f(x) . Since $(g \circ f )(x) = g \left [ f(x)\right ]$ , this is equivalent to evaluating g(x) for each value in the range of f(x) , as follows:

g(x) = 3x- 5

Range value: 3

g(3) = 3 (3)- 5 = 9 - 5 = 4

Range value: 5

g(5) = 3 (5)- 5 = 15 - 5 = 10

Range value: 8

g(8) = 3 (8)- 5 = 24 - 5 = 19

Range value: 13

g(13) = 3 (13)- 5 = 39 - 5 = 34

Range value: 21

g(21) = 3 (21)- 5 = 63 - 5 = 58

The range of g(x) on the set of range values of f(x) - and consequently, the range of $\left (g \circ f \right ) (x)$ - is the set $\left \{4 , 10, 19, 34, 58\right \}$ . Of the five choices, only 45 does not appear in this set; this is the correct choice.

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Example Question #1 : Miscellaneous Functions

Evaluate: $\sum_{k =2}^{5} (-1)^{k}(3k+1)$

Possible Answers:

Correct answer:

Explanation:

Evaluate the expression $(-1)^{k}(3k+1)$ for k = 2, 3, 4, 5 , then add the four numbers:

$a_{k} = (-1)^{k}(3k+1)$

$a_{2} = (-1)^{2}(3 \cdot 2 +1) = 1 (6+ 1) = 7$

$a_{3} = (-1)^{3}(3 \cdot 3 +1) = -1 (9+ 1) = -10$

$a_{4} = (-1)^{4}(3 \cdot 4 +1) = 1 (12+ 1) = 13$

$a_{5} = (-1)^{5}(3 \cdot 5 +1) =- 1 (15+ 1) = -16$

$\sum_{k =2}^{5} (-1)^{k}(3k+1) = 7 +( -10) + 13 + (-16 ) = -6$

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Example Question #1 : Miscellaneous Functions

Evaluate: $\sum_{k =3}^{7} (-2)^{k}$

Possible Answers:

248

Correct answer:

Explanation:

Evaluate the expression $\sum_{k =3}^{7} (-2)^{k}$ for k = 3, 4, 5, 6, 7 , then add the five numbers:

$a_{n} = (-2)^{k}$

$a_{3} = (-2)^{3} = -8$

$a_{4} = (-2)^{4 } =16$

$a_{5} = (-2)^{5 } =-32$

$a_{6} = (-2)^{6 } =64$

$a_{7} = (-2)^{7 } =-128$

$\sum_{k =3}^{7} (-2)^{k} = -8 + 16 + (-32)+ 64 +( -128 ) = -88$

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Example Question #1 : Miscellaneous Functions

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor$ and $g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

Which of the following is equal to $(f \circ g ) \left ( \frac{11}{4} \right )$ ?

Possible Answers:

Correct answer:

Explanation:

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right )$ , so, first, evaluate $g \left ( \frac{11}{4} \right )$ by substitution:

$g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

$g \left ( \frac{11}{4} \right ) = \left \lceil \frac{2}{3} \left ( \frac{11}{4} \right ) - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{11}{6} - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{4}{3} \right \rceil$

= 2

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right ) = f(2)$ , so evaluate f(2) by substitution.

$f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor$

$f(2) = \left \lfloor \frac{1}{2} (2)- \frac{2}{3} \right \rfloor$

$= \left \lfloor1 - \frac{2}{3} \right \rfloor$

$= \left \lfloor \frac{1}{3} \right \rfloor$

= 0 ,

the correct response.

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Example Question #1 : Miscellaneous Functions

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lceil 2.2x +6.3 \right \rceil$ and $g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$ .

Evaluate $(f \circ g )(3.4 )$

Possible Answers:

Correct answer:

Explanation:

$(f \circ g )(3.4 ) = f(g(3.4))$ , so first, evaluate g(3.4 ) using substitution:

$g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$

$g (3.4) = \left \lfloor -2.1 (3.4) + 9.3 \right \rfloor$

$= \left \lfloor -7.14+ 9.3 \right \rfloor$

$= \left \lfloor 2.16 \right \rfloor$

= 2

$(f \circ g )(3.4 ) = f(g(3.4)) = f(2)$ , so evaluate f(2) using substitution:

$f(x) = \left \lceil 2.2x +6.3 \right \rceil$

$= \left \lceil 2.2 (2) +6.3 \right \rceil$

$= \left \lceil 4.4 +6.3 \right \rceil$

$= \left \lceil 10.7 \right \rceil$

= 11 ,

the correct response.

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Example Question #1 : Miscellaneous Functions

Consider the polynomial

$2x^{5} + 3 x^{2}- 4x + k$ ,

where is a real constant. For x- 2 to be a zero of this polynomial, what must be?

Possible Answers:

k= 44

k= 68

k = -68

None of the other choices gives the correct response.

k= -44

Correct answer:

k = -68

Explanation:

By the Factor Theorem, x- c is a zero of a polynomial P(x) if and only if P(c) = 0 . Here, c = 2 , so evaluate the polynomial, in terms of , for x=2 by substituting 2 for :

$2 \cdot 2^{5} + 3 \cdot 2^{2}- 4 \cdot 2 + k$

$= 2 \cdot 32 + 3 \cdot 4 - 4 \cdot 2 + k$

=64 + 12 - 8 + k

=68 + k

Set this equal to 0:

68 + k = 0

k = -68

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College Algebra : Miscellaneous Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Miscellaneous Functions

Example Question #2 : Miscellaneous Functions

Example Question #1 : Miscellaneous Functions

Example Question #171 : College Algebra

Example Question #1 : Miscellaneous Functions

Example Question #1 : Miscellaneous Functions

Example Question #1 : Miscellaneous Functions

Example Question #1 : Miscellaneous Functions

Example Question #1 : Miscellaneous Functions

All College Algebra Resources

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