Determine algebraically whether the function is even, odd, or neither.

Question

Determine algebraically whether the function is even, odd, or neither. (show all work for full credit!)

 

f(x) = x^3/x^2+1

Answer 1

In order to test whether a function is even or odd, we do not look at the graph at all, we just plug in -x and see what comes back out. There are only three possible outcomes. If f(-x) turns out to be exactly f(x), the function is even. If f(-x) turns out to be -f(x), meaning every sign flipped, it is odd. And if it is neither of those, then the function is neither.

So let us take f(x) = x^3/(x^2 + 1) and replace every x with -x:

f(-x) = (-x)^3/((-x)^2 + 1)

Now we simplify the top and the bottom separately, and this is where students usually slip. On top, (-x)^3 means (-x)(-x)(-x). Two negatives make a positive and then the third one makes it negative again, so (-x)^3 = -x^3. On the bottom, (-x)^2 means (-x)(-x), and those two negatives cancel, so (-x)^2 = x^2. The odd exponent keeps the negative and the even exponent kills it:

f(-x) = -x^3/(x^2 + 1)

Remember, our purpose is to compare this to the original f(x) = x^3/(x^2 + 1). The bottom is identical and the top is the exact opposite, so the whole fraction is the opposite of what we started with. We can factor that negative out front:

f(-x) = -(x^3/(x^2 + 1)) = -f(x)

That matches the definition of an odd function.

So, our final answer is that f(x) = x^3/(x^2 + 1) is odd.

Answer 2

A function is even if f(-x) = f(x).  A function is odd if f(-x) = -f(x).

 

f(x) = x³ / (x²+1)

 

f(-x) = (-x)³ / ( (-x)²+1) = -x³ / (x²+1)

 

                                     = -f(x)

 

So, the given function is an ODD function.