Determining Even and Odd Functions

When we think “even and odd,” usually even and odd numbers are what come to mind. But what are even and odd functions? In today’s video, we will define even and odd functions and discuss how to identify them.

Let’s begin by talking about even functions. If a function \(f(x)\) evaluated at \(-x\) gives us the same \(f(x)\) we started with, that function is even. Formally written, when

\(f(x)=f(-x)\)

\(f(x)\) is even. Graphically, an even function will appear symmetric, or mirrored, about the vertical y-axis. One example of an even function is \(f(x)=x^{2}\).

f(x)=x^2 on a graph. The sides of the curve are identical.

Notice that the shape of this familiar parabola is visibly symmetric. The left and right sides of the plane are identical, just flipped. We can also show that this function is even algebraically, by evaluating at \(-x\).

So our original function is \(f(x)=x^{2}\). And we said that, if \(f(-x)\) is the same as \(f(x)\), then the function is even. So let’s evaluate at \(-x\). So, wherever there’s an \(x\), we’re gonna plug in \(-x\). So we have:

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

Which, when you square a negative, it turns positive, so this is equal to:

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

So in this case, \(f(-x)=f(x)\). And because of our definition of even, this function, \(f(x)=x^{2}\), is even.

Notice that if we add a constant to this function, it won’t affect the shape of the function, just raise or lower it on the plane. For example, this is the graph of \(f(x)=x^{2}+1\). Notice how it still has the same shape, and it is still an even function. It has just been moved up one unit on the coordinate plane.

Image of f(x)=x2+1 written on a graph. The point is raised.

Now let’s talk about what odd functions are like. Consider another function \(f(x)\), which we will once again evaluate at \(-x\). But this time, instead of looking for the same \(f(x)\) we started with, we want to see if \(f(-x)\) changes the sign of all terms in the function. In other words, if \(f(-x)=-f(x)\), then the function is odd.

Graphically, an odd function will appear the same when we rotate it by 180°, like flipping a page upside down, and it must pass through the origin. A few examples of odd functions are:

\(f(x)=x\),

\(f(x)=x^{3}\),

and \(f(x)=sin(x)\)

Let’s take a look at what’s going on here algebraically, using \(f(x)=x^{3}\) as an example.

So we’re gonna have our original function: \(f(x)=x^{3}\). And just like before, we’re gonna evaluate it at \(-x\). So we want \(f(-x)\). So anywhere we see an \(x\), we’re gonna plug in \(-x\).

\(f(-x)=(-x)^{3}\)

So \((-x)^{3}\), which is \(-x\cdot-x\cdot-x\), which means, since there are three negatives, our final answer’s gonna be \(-x^{3}\).

\(f(-x)=-x^{3}\)

So if you notice, \(f(x)\) is the opposite of \(f(-x)\). Each term, which in this case we only have one, is changed from positive to negative. So that means that this function is odd.

Notice that if we were to add a constant to this function, it would no longer be odd. Remember that for odd functions, every term must change signs when evaluating at \(-x\). That constant term would have no way to change sign, and we would see on the graph that the function would no longer pass through the origin.

So we have now talked about definitions of both even and odd functions but before we go further, it’s important to clarify that some functions may be neither even nor odd! For example, take a look at the function \(f(x)=(x+1)^{2}\).

Image demonstrating that some graphs have equations that are neither even nor odd.

From the graph we can see that this function doesn’t pass through the origin, so it can’t be odd. And it isn’t symmetric about the \(y\)-axis, so it isn’t even either. But we can also determine this algebraically.

So our function is \(f(x)=(x+1)^{2}\). And remember, to determine if it’s even or odd, we want to evaluate it at \(-x\). So we have:

\(f(-x)=(-x+1)^{2}\)

Which we can write as:

\(f(-x)=(-x+1)^{2}=(-x+1)(-x+1)\)

‘cause that’s what the squared means. And then from here we can FOIL.
So:

\(-x\cdot-x=x^{2}\) 
 \(1\cdot-x=-x\) 
 \(-x\cdot1=-x\)

And,

\(1\cdot1=1\)

So we have:

\(f(-x)=x^{2}-2x+1\)

But remember, to determine even or odd, we have to compare this to our original function. So let’s expand that. So for this, we have:

\(f(x)=(x+1)(x+1)\)

Which is:

\(x⋅x=x^{2}\) 

 \(1\cdot x=x\text{ plus }x\cdot1=x,\text{ so plus }2x\) 
 
\(1\cdot1=1\) 
 
\(f(x)=(x+1)(x+1)=x^{2}+2x+1\)

Now, when we compare these two functions, we see that only one of the three terms ended up changing signs, so \(f(x)\) is not odd. And since one term did change sign, \(f(x)\neq f(-x)\), so the function is not even either.

Now that we’ve laid a groundwork for understanding even and odd functions, let’s talk about why we call them even and odd in the first place.

Remember that even functions are the same when we evaluate them at \(+x\) and at \(-x\). As we saw earlier, \(f(x)=x^{2}\) satisfies this property because anytime we square something, a positive value is returned, and therefore the sign of that term doesn’t change even if we plug a negative value in. The same is true when something is raised to the fourth power, or the sixth, and so on. Notice that constants do not change sign when we evaluate at \(-x\) either. That’s why we saw that the function \(f(x)=x^{2}+1\) was still even. As you can see, an even function will have even exponents.

It may be unsurprising now that odd functions likewise will have odd exponents! Remember that in order for a function to be odd, all terms must change sign when we evaluate at \(-x\).

Clearly, any term with \(x\) to the first power will change sign when we plug in a negative value of \(x\). In the same way, \(x\) to the third power, the fifth power, and so on will all change sign when we plug in a negative value for \(x\). As we mentioned earlier, when a term has an even power of \(x\), it will not change sign. That means that an odd function cannot contain any terms with even powers of \(x\), and it cannot have any constants.

You may recognize even and odd functions later on in calculus when it comes to dealing with the Taylor expansion.

Time for some practice problems!

Based on this graph, is this function even, odd, or neither?

A graph that is not odd or even.

Neither. This function is not symmetric about the \(y\)-axis, so it is not even. And even though it passes through the origin, it is not odd either because it would not appear the same if we were to rotate the image 180°.

Let’s try another one.

Is the function \(f(x)=\frac{5}{7}x^{3}-2x\) even, odd, or neither?

Let’s look at each term. First, \(\frac{5}{7}x^{3}\) has an odd power of \(x\), meaning that the sign will change when evaluated at \(-x\). Similarly, the second term, \(-2x\), has an odd power of \(x\) and will also change sign. That means this function is odd!

Let’s finish with a more conceptual question.

We know that some functions may be neither even nor odd, but is it possible for a function to be both even and odd?

Surprisingly, the answer is yes, but only for one function. Can you think of what function that is? Remember that for even functions, \(f(-x)=f(x)\), and for odd functions, \(f(-x)=-f(x)\). The only way both of these can be satisfied is when \(f(x)=0\).

\(f(-x)=f(x)\) 
 \(and\) 
 \(f(-x)=-f(x)\)

As a quick recap, we can identify even and odd functions in the following ways:

Graphically, even functions are symmetric about the \(y\)-axis. And they don’t have to pass through the origin. Though, odd functions must pass through the origin, and they will appear the same when viewed from a 180° rotation.

Algebraically, even functions are the same when we evaluate at \(+x\) and at \(-x\). Odd functions will change signs across all terms when evaluated at \(-x\).

As a shortcut, if a function contains only even exponents of \(x\) (and may or may not have constants) then it is even. If a function has no constants and only odd exponents of \(x\), then it is odd.

Now that we’ve covered everything and run through some examples, you should be pretty comfortable with identifying even and odd functions.

Thanks for watching, and happy studying!

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by Mometrix Test Preparation | Last Updated: August 25, 2021