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College Algebra : College Algebra

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Example Questions

Example Question #1 : Symmetry

Relation

Which of the following is true of the relation graphed above?

Possible Answers:

It is not a function

It is a function, but it is neither even nor odd.

It is an odd function

It is an even function

Correct answer:

It is an odd function

Explanation:

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Relation

Also, it is seen to be symmetric about the origin. Consequently, for each in the domain, f(-c) = -f(c) - the function is odd.

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Example Question #2 : Symmetry

f(x) is an even function; f(9) = 10 .

True or false: It follows that f(10) = 9 .

Possible Answers:

True

False

Correct answer:

False

Explanation:

A function f(x) is even if and only if, for all in its domain, f(-x) = f(x) . It follows that if f(9) = 10 , then

f(-9)= f(9) = 10 .

No restriction is placed on any other value as a result of this information, so the answer is false.

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Example Question #3 : Symmetry

Relation

The above table refers to a function f(x) with domain $\left \{ -3, -2, -1, 0, 1, 2, 3 \right \}$ .

Is this function even, odd, or neither?

Possible Answers:

Odd

Neither

Even

Correct answer:

Neither

Explanation:

A function f(x) is odd if and only if, for every in its domain, f(-c) = -f(c) ; it is even if and only if, for every in its domain, f(-c) = f(c) .

We see that f(-3) = -8 and f(3) = 8 . Therefore, $f(-3)\ne f(3)$ , so f(-c) = f(c) is false for at least one . f(x) cannot be even.

For a function to be odd, since f(-0) = - f(0) , it follows that f(0) = - f(0) ; since f(0) is its own opposite, f(0) must be 0. However, f(0) = 7 ; f(x) cannot be odd.

The correct choice is neither.

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Example Question #71 : Graphs

Define $f(x) = \frac{10}{x- 1}$ .

Is this function even, odd, or neither?

Possible Answers:

Even

Odd

Neither

Correct answer:

Neither

Explanation:

A function f(x) is odd if and only if, for all , f(-x) = -f(x) ; it is even if and only if, for all , f(-x) = f(x) . Therefore, to answer this question, determine f(-x) by substituting for , and compare it to both f(x) and - f(x) .

$f(x) = \frac{10}{x- 1}$

$f(-x) = \frac{10}{-x- 1}$

$f(-x) \ne f(x)$ , so f(x) is not even.

$- f(x) = - \frac{10}{x- 1} = \frac{10}{-x+1}$

$f(-x) \ne - f(x)$ , so f(x) is not odd.

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Example Question #6 : Symmetry

f(x) is a piecewise-defined function. Its definition is partially given below:

$f(x) = \left\{\begin{matrix}? &x < 0 \\x^{4}+ x^{3}, & x \ge 0 \end{matrix}\right.$

How can f(x) be defined for negative values of so that f(x) is an odd function?

Possible Answers:

$f(x) = x^{4}-x^{3}$

$f(x) = - x^{4}-x^{3}$

$f(x) = x^{4}+x^{3}$

f(x) cannot be made odd.

$f(x) = - x^{4}+x^{3}$

Correct answer:

$f(x) = - x^{4}+x^{3}$

Explanation:

f(x) , by definition, is an odd function if, for all in its domain,

f(-x) = -f(x) , or, equivalently f(x) = -f(-x)

One implication of this is that for to be odd, it must hold that f(0) = 0 . If x= 0 , then, since

$f(x)= x^{4}+ x^{3}$

for nonnegative values, then, by substitution,

$f(0)= 0^{4}+ 0^{3} = 0$

This condition is satisfied.

Now, if is negative, is positive. it must hold that

f(x) = -f(-x) ,

so for all x < 0

f(x) = -f(-x)

$f(x) = -[(-x)^{4}+ (-x)^{3}]$

$f(x) = -( x^{4}-x^{3} )$

$f(x) = - x^{4}+x^{3}$ ,

the correct response.

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Example Question #3 : Symmetry

Function 3

Consider the relation graphed above. Which is true of this relation?

Possible Answers:

The relation is a function which is neither even nor odd.

The relation not a function.

The relation is an even function.

The relation is an odd function.

Correct answer:

The relation is a function which is neither even nor odd.

Explanation:

The relation passes the Vertical Line test, as seen in the diagram below, in that no vertical line can be drawn that intersects the graph than once:

Function 3

An function is odd if and only if its graph is symmetric about the origin, and even if and only if its graph is symmetric about the -axis. From the diagram, we see neither is the case.

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Example Question #8 : Symmetry

f(x) is a piecewise-defined function. Its definition is partially given below:

$f(x) = \left\{\begin{matrix}? &x < 0 \\ 0, & x = 0 \\ \ln (x^2+x), & x > 0 \end{matrix}\right.$

How can f(x) be defined for negative values of so that f(x) is an odd function?

Possible Answers:

$f(x) = \ln (-x^{2}+x )$

$f(x) = - \ln (-x^{2}+x )$

$f(x) = \ln (x^{2}-x )$

$f(x) = - \ln (x^{2}+x )$

$f(x) = - \ln (x^{2}-x )$

Correct answer:

$f(x) = - \ln (x^{2}-x )$

Explanation:

f(x) , by definition, is an odd function if, for all in its domain,

f(-x) = -f(x) , or, equivalently f(x) = -f(-x)

One implication of this is that for to be odd, it must hold that f(0) = 0 . Since f(0) is explicitly defined to be equal to 0 here, this condition is satisfied.

Now, if is negative, is positive. it must hold that

f(x) = -f(-x) ,

so for all x < 0

$f(x) = - \ln[ (-x)^2+(-x)]$

$f(x) = - \ln (x^{2}-x )$

This is the correct choice.

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Example Question #4 : Symmetry

Which of the following is symmetrical to y=-2x+3 across the origin?

Possible Answers:

$y= -2x+\frac{3}{2}$

$y= -\frac{1}{2}x+\frac{3}{2}$

$y= -\frac{1}{2}x-\frac{3}{2}$

$y= \frac{1}{2}x-\frac{3}{2}$

$y= \frac{1}{2}x+\frac{3}{2}$

Correct answer:

$y= -\frac{1}{2}x+\frac{3}{2}$

Explanation:

Symmetry across the origin is symmetry across y=x .

Determine the inverse of the function. Swap the x and y variables, and solve for y.

x=-2y+3

Subtract 3 on both sides.

x-3=-2y+3-3

x-3=-2y

Divide by negative two on both sides.

$\frac{x-3}{-2}=\frac{-2y}{-2}$

The answer is: $y= -\frac{1}{2}x+\frac{3}{2}$

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Example Question #71 : Graphs

Consider the function $f(x)= x^{8}+ 2x^{4}+ 1$ .

Is an even function, an odd function, or neither?

Possible Answers:

Odd

Neither

Even

Correct answer:

Even

Explanation:

A function is even if, for each in its domain,

f(-c)= f(c) .

It is odd if, for each in its domain,

f(-c)=- f(c) .

Substitute for in the definition:

$f(-x)= (-x)^{8}+ 2(-x)^{4}+ 1$

$= (-1)^{8}x^{8}+ 2 (-1)^{4}x^{4}+ 1$

$= 1x^{8}+ 2 (1) x^{4}+ 1$

$= x^{8}+ 2x^{4}+ 1$

= f(x)

Since f(x) = f(-x) , is an even function.

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Example Question #11 : Symmetry

f(x) is an even function. Let $g(x) = \frac{f(x)}{x}$ .

Is an even function, an odd function, or neither?

Possible Answers:

Odd

Neither

Even

Correct answer:

Odd

Explanation:

A function is even if, for each in its domain,

f(-c)= f(c) .

It is odd if, for each in its domain,

f(-c)=- f(c) .

Substitute for in the definition of :

$g(-x) = \frac{f(-x)}{-x} = - \frac{f(-x)}{x}$

Since is even, f(-x)= f(x) , so

$g(-x) = - \frac{f(x)}{x} = -g(x )$

This makes an odd function.

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College Algebra : College Algebra

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Symmetry

Example Question #2 : Symmetry

Example Question #3 : Symmetry

Example Question #71 : Graphs

Example Question #6 : Symmetry

Example Question #3 : Symmetry

Example Question #8 : Symmetry

Example Question #4 : Symmetry

Example Question #71 : Graphs

Example Question #11 : Symmetry

All College Algebra Resources

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