Help with a calculus problem?

Question

Discuss whether each of the following functions is odd, even, or neither. Use algebraic as well as graphical reasoning

f(x)=cos2x
y=(x)/(x^(2)+1)
y=x^(2/3)

Answer 1

Recall the following definitions:
A function is even if and only if f(x) = f(-x).
A function is odd if and only if f(-x) = -f(x).

1) f(x) = cos(2x)

To test if this functions is even, we calculate f(-x).
f(-x) = cos(-2x)

We can rewrite this.

cos(-2x) = cos(0 - 2x)

Now, we use the cosine subtraction identity. It states that

cos(0 - 2x) = cos(0)cos(2x) + sin(0)sin(2x)
= (1)cos(2x) + 0sin(2x)
= cos(2x), which is equal to f(x).
Therefore, f(x) = f(-x), and this function is even.


2) y = x/[x^2 + 1]

Let y = f(x). Then
f(x) = x/[x^2 + 1]

f(-x) = -x/[ (-x)^2 + 1 ] = -x/[x^2 + 1], which is clearly not equal to f(x). Therefore, this function is not even.

Now, we compare f(-x) to -f(x).
-f(x) = -[x/[x^2 + 1]] = -x/(x^2 + 1). As you can see, this function is equal to f(-x) since we just calculated that value above.

Therefore, this function is odd.

3) y = x^(2/3).
Let f(x) = x^(2/3)

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

Since -1 to the power of 2/3 is equivalent to taking the cube root of -1 and then squaring, we get 1. So
f(-x) = x^(2/3), which is clearly equal to f(x).

Therefore, this function is even.

Answer 2

f(x)=cos(2x) is even
the next function is odd
the last function is even