What does $g(x) = \frac{1}{2} (f(x) + f(-x))$ mean?

Question

Suppose any function $f(x)$ defines $g(x)=\frac{1}{2} (f(x)+f(-x))$. What does this mean essentially? The question is to show $g(x)$ is even, so I am doing $$g(-x) = g(x) \Rightarrow g(-x)= \frac{1}{2} (f(-x)+f(-(-x))).$$ However, I don't understand why the $x$ in the $f(x)$ function inside the $g(x)$ also has to be $-x$ when I choose to do $g(-x)$, I am doing it just because and it works but I don't know why. What does it mean for a function to be defined in terms of another function?

Answer 1

The function is NOT $f(x)$, the same way $f(a)$ is not a function. The function is simply $f$. A function $$ f: A \rightarrow B $$ has a domain $A$ and a co-domain $B$. When we write $f(a)$, for some $a \in A$, we are talking about a value in $B$ which is associated to $a \in A$.

But how can I tell someone exactly what the function is? How to determine $f(a)$ from $a$? One way to do that is to give a formula. For example: $$ \begin{align*} f: \mathbb{R} &\rightarrow \mathbb{R} \\ x &\mapsto 2x. \end{align*} $$ The function described above has NOTHING to do with the variable $x$. It could as well be written as $$ \begin{align*} f: \mathbb{R} &\rightarrow \mathbb{R} \\ a &\mapsto 2a. \end{align*} $$ Both are the same, with $x$ or with $a$. What both formula say is that to calculate $f$, for a given $k \in A$, all you have to do is multiply it by $2$. That is, the function $f$ takes its input, doubles its value and present the result as its output.

If we are to write a formula for the function given its input, we usually name the input $x$ and express how to calculate the output based on $x$. If you want to make it less confuse, you can, for example, rewrite $g$'s definition to read $$ g(y) = \frac{1}{2}(f(y) + f(-y)). $$ What it says is that

$g$ takes some value, puts this value and also minus its value into $f$, adds the results and divide by $2$.

Those words would be easier if we give a name to "some value". We called it $y$, but we could have called it $x$... and this $x$ has nothing to do with any other $x$ you might be using elsewhere. We just happened to have labeled this "variable" $x$ so we can express $f$'s formula.

So, to calculate $g(-x)$, what you do is substitute $y$ above by $-x$: $$ g(-x) = \frac{1}{2}(f(-x) + f(-(-x))). $$

Answer 2

Take $f(x) = x + 1$, for example. Then $$g(x) = \frac{1}{2}\big(f(x)+f(-x)\big)$$ means that $g(x) = \frac{1}{2}\big((x+1) + (-x+1)\big) = 1$ (even function).

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If a function is defined in terms of another function, say, $g(x) = F(f(x))$ for some rule (function) $F$, then we can find any value of $g$ by plugging the corresponding values of $f$ into the rule $F$. In the above example, $f(x) = x + 1$ and $F_x = \frac{1}{2}(f_x + f_{-x})$.

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The following chain can describe the process. $$x \overset f\to y \overset F\to z$$ We start by taking some value $x$, which obeys the rules. The first rule $f$ orders $x$ to go to $y$, then the rule $F$ sends $y$ to $z$, which is the final value obtained by two rules $F(f(x))) = g(x)$.

Answer 3

When you have any function say $g(x)$,

you get $g(-x)$ by substituting in (-x) whereever x is in g(x). This is just by definition. Similarly if you were trying to get $g(y^2+2)$, it's the same idea. Replace every occurence of x in $g(x)$ with $(y^2+2)$

So if $g(x) = -3x^3-4x^2+6x+2$

To get $g(-x)$, you replace every appearance of x with (-x)

So $g(-x)$ in this case is $-3(-x)^3-4(-x)^2+6(-x)+2$

and

$g(y^2+2) = -3(y^2+2)^3-4(y^2+2)^2+6(y^2+2)+2$

In your case $g(x) = 0.5(f(x)+f(-x))$

What is $g(-x)$ in this case?

Replace the occurence of x with (-x) everywhere... including the spots inside other functions like f.

$g(-x) = 0.5(f((-x)) + f(-(-x)))$

I can replace $f((-x))$ by $f(-x)$ because $(-x) = -x$. I can simplify $f(-(-x))$ by $f(x)$ because $-(-x) = x$

$g(-x) = 0.5(f(-x) + f(x))$

$g(-x) = 0.5(f(x) + f(-x))$

So since we can see that $g(-x) = g(x)$, by definition this $g(x)$ is even.

As André Caldas describes the function g takes a 'something' and does certain operations on that something... The best way to illustrate this is using a dummy variable like x and writing out $g(x)$. g(something else) replaces occurence of x with (something else)