Why must we account for the domain of an original function when we reduce?

Question

Why are reduced rational functions not always "equal to" the original function? Why must we account for the domain of an original function when we reduce? For example,

$$g(x) = \dfrac{x^2 + 3x - 4}{x - 1}$$

has a limited domain, because $x = 1$ will cause division by zero. Thus, there will be a hole in the graph at $x = 1$. However, the reduced function,

$$f(x) = x + 4$$

does not have a limited domain, and therefore will not have a hole, but we are supposed to put one in anyway.

I can't really understand why this is the case. It sort of makes sense, but I would appreciate an explanation for it. I tried to rationalize it in my brain by thinking of the following situation:

If I have 8 one-quarter pieces of pie, I have two whole pies because $ 8 \times (1/4) = 2$. However, it is not the same as having two whole pies because they have slices in them.

Is this an "okay" analogy to think about what is going on here?

Answer 1

Since one of them has a hole and the other doesn't, they're not equal. If $x\ne1$, then $f(x)=g(x)$. If $x=1$, then $f(x)=f(1)=5$ and $g(x)$ reduces to $0/0$, so it's undefined. That's why the reduced rational function is not equal to the original rational function.

You say the reduced function does not have a hole. That is precisely why it is not equal to the original function. But if you "put one in anyway", then what you get is equal to the original function.

One of the most important reasons for reducing is precisely to remove the hole.

If a car goes $60$ miles in $2$ hours, then dividing, you conclude that its average speed during that time is $60/2=30$ miles per hour.

If the car goes $0$ miles in $0$ hours, what is its speed at that time? Is it $0/0$ miles per hour? No matter what the car's speed is at that instant, you get $0/0$. But reducing and thereby getting rid of the hole is the way to find the speed at an instant.

Answer 2

First write the function as $$f(x) = \frac{(x+4)(x-1)}{x-1}.$$ What you would like to do at this point is cancel the factor $x - 1$ from the numerator and denominator. What does this actually mean though? It means that we divide the numerator by $x - 1$ and we divide the denominator by $x - 1$; as we have done the same thing to both parts of the fraction, what remains is equivalent to the original expression. When can we divide by $x-1$? Precisely when $x-1\neq 0$, i.e. $x\neq 1$. Therefore, for $x \neq 1$, $f(x) = x + 4$.