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A level - algebra and functions

I am given that g(x) = x+1 / X-2 and the domain: x > 3
The answer says that the range is 1 < y < 4, how do I get this answer without drawing a graph?
The explanation is that g(x) is decreasing in the domain, and "approaches and bounded below by 1 as x gets really big" (another way of saying the horizontal asymptote is y=1). Then you'll get the range.

But you'll likely never figure this out without drawing a graph. The shortcut to finding range really is to draw the graph.
(edited 1 month ago)
Original post by tonyiptony
The explanation is that g(x) is decreasing in the domain, and "approaches and bounded below by 1 as x gets really big" (another way of saying the horizontal asymptote is y=1). Then you'll get the range.
But you'll likely never figure this out without drawing a graph. The shortcut to finding range really is to draw the graph.

thank you☺️
Reply 3
Original post by helenastathakis
I am given that g(x) = x+1 / X-2 and the domain: x > 3
The answer says that the range is 1 < y < 4, how do I get this answer without drawing a graph?

You can also write it as
((x-2) + 3) / (x-2) = 1 + 3/(x-2)
so make it proper. Then its a bit easier to see it tends to 1 for large x and equals 4 when x=3.
Original post by helenastathakis
I am given that g(x) = x+1 / X-2 and the domain: x > 3
The answer says that the range is 1 < y < 4, how do I get this answer without drawing a graph?

Just to add to @tonyiptony 's comments, with functions of this kind it is often helpful to attempt an algebraic rearrangement. By algebraic division, (x + 1) / (x - 2) can be expressed as 1 + 3/(x - 2). The second part of that expression can be recognised as a linear transformation of 1/x, and this may aid visualisation of the shape of the curve as a whole.
(edited 1 month ago)
I often trust my curve sketching skills more than algebraic manipulation, so your choice really.
If you want to solve it by observation (which I tend to lean towards):
In case of functions where there is a variable in denominator, see at what value of y will there be no x left in the equation. In this case, y=1, because substituting it will lead to the removal of x from the equation, hence no sol.
For the upper bound, I would substitute the domain. y=4 when x=3. Because the domain states that x>3, then y<4.
So you get 1<y<4.
The methods may vary (slightly) with the type of equation but observation takes up a lot less time, no?

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