Miscellaneous Functions - College Algebra

Card 0 of 9

Question

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

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

Answer

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 . Since $(g \circ f )(x) = g \left [ f(x)\right ]$ , this is equivalent to evaluating for each value in the range of , as follows:

Range value: 3

Range value: 5

Range value: 8

Range value: 13

Range value: 21

The range of on the set of range values of - 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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Question

Refer to the above diagram, which shows the graph of a function .

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

Answer

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

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

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Question

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

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

Answer

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

Replace with :

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

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

Switch the positions of and :

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

or

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

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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Question

The above diagram shows the graph of function on the coordinate axes. True or false: The -intercept of the graph is

Answer

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:

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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Question

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

Answer

Evaluate the expression $(-1)^{k}(3k+1)$ for , 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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Question

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

Answer

Evaluate the expression $\sum_{k =3}^{7} (-2)^{k}$ for , 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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Question

Consider the polynomial

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

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

Answer

By the Factor Theorem, is a zero of a polynomial if and only if . Here, , so evaluate the polynomial, in terms of , for by substituting 2 for :

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

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

Set this equal to 0:

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Question

$\left \lfloor N \right \rfloor$ 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 $(f \circ g ) \left ( \frac{11}{4} \right )$ ?

Answer

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

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

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

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

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

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

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

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

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

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

,

the correct response.

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Question

$\left \lfloor N \right \rfloor$ 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 $(f \circ g )(3.4 )$

Answer

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

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

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

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

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

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

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

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

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

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

,

the correct response.

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