Question

Intuition for why solving for x works for algebraically finding range of a function?

Answer 1

The range is the set of possible y values. A y is in the range if there's some x that can produce it.

So you're finding an explicit rule for the x that can produce a given y. If you have such a rule, you know the y's that can be produced.

Let's take a simple example, like y = x\^2 + 6, a parabola that we know starts at y = 6 and goes upward. We'll restrict it to x >= 0 so it's invertible.

What x produces a given y? Well inverting it, we get x = sqrt(y - 6). If y >= 6, we can use this formula to explicitly find the x that produces that y, that has the property that y = x\^2 + 6. Since that formula produces an answer for any y >= 6, we know that all y >= 6 are in the range.

But if y < 6, the formula fails. There is no x such that x = sqrt(y - 6). And that means there's no x such that y = x\^2 + 6 that will produce a y < 6. So the values < 6 are not in the range.

Answer 2

It doesn't. Take arctangent for example. You can only count on the trick for invertible functions. (If you're instructed to use it, go ahead, because instructions.)

That said, you basically answered your own question:

>finding restrictions on y

Range means asking what are the restrictions on y. Mezzo pretty much explained it.

Answer 3

Suppose you want to find the range of y=x^(2)-4.

You can solve for x, x=sqrt(y+4), and see what values of y will give reasonable answers; y has to be at least -4, which is the range of the original function.

It's just an alternative way to look at functions where this type of solution is feasible.