Is it still an even function if I replace $y$ with $-y$

Question

I have a function, say, $x=y^2$ which is of the form $x=f(y)$ so if I replace $y$ with $-y$ and I get $f(y) = f(-y)$ can I still call it an Even function?

The reason I'm asking this is because everywhere I see, the definition is "if you replace $x$ with $-x$ and you get $f(x) = f(-x)$ then it's an even function" instead of "if you replace the independent variable of the function with it's additive inverse and the function remains unchanged then it's an even function"

I feel the former definition is a bit more specific, hence this question arises.

Answer 1

What you can say is that $x$ is an even function of $y$

If $f(x)=f(-x)$ then $f(x)$ is an even function of $x$. It doesn't matter that the letter is $x$, but it does matter what the independent variable is.

For example, suppose $f(x)=x^4+x^2$, this is an even function of $x$.

But if I put $z=x^2$ then $g(z)=z^2+z$ is not an even function of $z$ (even though $g(x^2)=f(x)$)

Answer 2

A function $f$ is said even when

$$\forall x, f(x)=f(-x).$$

There is no ambiguity in this statement.

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$$\forall x, x^2=(-x)^2$$

is true, so the square function is even.

$$\forall x, x+1=-x+1$$

is false, so the "plus one"function is not even.