Symmetry - College Algebra

, by definition, is an odd function if, for all in its domain,

, or, equivalently

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

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

,

so for all

This is the correct choice.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

One implication of this is that for to be odd, it must hold that . If , then, since

for nonnegative values, then, by substitution,

This condition is satisfied.

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

,

so for all

,

the correct response.

A function is odd if and only if, for every in its domain, ; it is even if and only if, for every in its domain, .

We see that and . Therefore, , so is false for at least one . cannot be even.

For a function to be odd, since , it follows that ; since is its own opposite, must be 0. However, ; cannot be odd.

The correct choice is neither.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

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

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

,

so for all

This is the correct choice.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

One implication of this is that for to be odd, it must hold that . If , then, since

for nonnegative values, then, by substitution,

This condition is satisfied.

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

,

so for all

,

the correct response.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

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

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

,

so for all

This is the correct choice.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

One implication of this is that for to be odd, it must hold that . If , then, since

for nonnegative values, then, by substitution,

This condition is satisfied.

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

,

so for all

,

the correct response.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

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

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

,

so for all

This is the correct choice.

, by definition, is an odd function if, for all in its domain,

, or, equivalently

One implication of this is that for to be odd, it must hold that . If , then, since

for nonnegative values, then, by substitution,

This condition is satisfied.

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

,

so for all

,

the correct response.

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:

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

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

, so is not even.

, so is not odd.

A function is even if and only if, for all in its domain, . It follows that if , then

.

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

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

.

It is odd if, for each in its domain,

.

Substitute for in the definition of :

Since is even, , so

This makes an odd function.

To check for symmetry, we are going to do three tests, which involve substitution. First one will be to check symmetry along the x-axis, replace .

This isn't equivilant to the first equation, so it's not symmetric along the x-axis.

Next is to substitute .

This is not the same, so it is not symmetric along the y-axis.

For the last test we will substitute , and

This isn't the same as the orginal equation, so it is not symmetric along the origin.

The answer is it is not symmetric along any axis.

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:

Also, it can be seen to be symmetrical about the origin. Consequently, for each in the domain, - the function is odd.

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:

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.

Symmetry across the origin is symmetry across .

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

Subtract 3 on both sides.

Divide by negative two on both sides.

The answer is:

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

.

It is odd if, for each in its domain,

.

Substitute for in the definition:

Since , is an even function.

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:

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