College Algebra : Miscellaneous Functions

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #81 : Graphs

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

Replace \displaystyle f(x) with \displaystyle y:

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

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

Switch the positions of \displaystyle y and \displaystyle x:

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

or

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

Solve for \displaystyle y. This can be done as follows:

Square both sides:

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

\displaystyle 4y- 9 = x ^{2}

Add 9 to both sides:

\displaystyle 4y- 9+9 = x ^{2}+9

\displaystyle 4y = x ^{2}+9

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

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

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

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

Replace \displaystyle y with \displaystyle f^{-1}(x):

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

Example Question #1 : Miscellaneous Functions

Function

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

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

Possible Answers:

True

False

Correct answer:

False

Explanation:

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

Function

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

Example Question #1 : Miscellaneous Functions

Function

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

The \displaystyle y-intercept of the graph of a function is the point at which it intersects the \displaystyle y-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 \displaystyle y-intercept \displaystyle \left ( 0, 1\frac{3}{4}\right ).

Example Question #3 : Miscellaneous Functions

Function

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

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

Possible Answers:

\displaystyle 58

\displaystyle 10

\displaystyle 34

\displaystyle 45

\displaystyle 19

Correct answer:

\displaystyle 45

Explanation:

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

 

\displaystyle g(x) = 3x- 5

Range value: 3 

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

Range value: 5

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

Range value: 8

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

Range value: 13

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

Range value: 21

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

 

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

Example Question #91 : Graphs

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

Possible Answers:

\displaystyle -46

\displaystyle 6

\displaystyle 0

\displaystyle -6

\displaystyle 46

Correct answer:

\displaystyle -6

Explanation:

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

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

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

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

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

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

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

Example Question #2 : Miscellaneous Functions

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

Possible Answers:

\displaystyle -248

\displaystyle -88

\displaystyle 248

\displaystyle 0

\displaystyle 88

Correct answer:

\displaystyle -88

Explanation:

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

\displaystyle a_{n} = (-2)^{k}

\displaystyle a_{3} = (-2)^{3} = -8

\displaystyle a_{4} = (-2)^{4 } =16

\displaystyle a_{5} = (-2)^{5 } =-32

\displaystyle a_{6} = (-2)^{6 } =64

\displaystyle a_{7} = (-2)^{7 } =-128

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

Example Question #2 : Miscellaneous Functions

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

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

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

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

Possible Answers:

\displaystyle -1

\displaystyle 0

\displaystyle -2

\displaystyle 2

\displaystyle 1

Correct answer:

\displaystyle 0

Explanation:

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

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

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

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

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

\displaystyle = 2

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

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

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

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

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

\displaystyle = 0,

the correct response.

Example Question #3 : Miscellaneous Functions

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

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

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

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

Possible Answers:

\displaystyle -21

\displaystyle 11

\displaystyle -19

\displaystyle 10

\displaystyle -20

Correct answer:

\displaystyle 11

Explanation:

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

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

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

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

\displaystyle = \left \lfloor 2.16 \right \rfloor

\displaystyle = 2

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

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

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

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

\displaystyle = \left \lceil 10.7 \right \rceil

\displaystyle = 11,

the correct response.

Example Question #172 : College Algebra

Consider the polynomial

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

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

Possible Answers:

None of the other choices gives the correct response.

\displaystyle k= -44

\displaystyle k= 68

\displaystyle k= 44

\displaystyle k = -68

Correct answer:

\displaystyle k = -68

Explanation:

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

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

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

\displaystyle =64 + 12 - 8 + k

\displaystyle =68 + k

Set this equal to 0:

\displaystyle 68 + k = 0

\displaystyle k = -68

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