How do i find the range of the function x+1/x-2 x>3
You should put brackets in because the expression you wrote could be a few things. If you mean (x+1)/(x-2) Think about what happens as x goes to infinity. Also you should be to see what the maximum vale of this function is just from putting in one or two values of x and seeing what the value is.
Yes, because nowhere after x>3 does the y value reach 0. As you can see from the graph (or calculate from using limits), as x tends to positive infinity, y tends to 1.
Yes, because nowhere after x>3 does the y value reach 0. As you can see from the graph (or calculate from using limits), as x tends to infinity, y tends to 1.
What happens at the beginning of the interval, when x is equal to 3? What happens when X is extremely large? When X is extremely large, does the +1 and the -2 matter? Could you possibly ignore the +1 and -2 to simplify the expression at larger values of x?
I did a visual representation, so you can see it graphically....
As you can see from the f(x) graph attached, you can see that when x=3, the y value must be equal to 4.
So, as the x value is greater then 3 for this graph, you sub in 4 etc.. and you get your y value...
The graph should be drawn from the straight line where x=3 (see attachment), but it will still tend to 0.
Therefore, the range will be that: (as seen on the graph)
4>y>0
It doesn't tend to Zero. That's wrong. At the beginning of the interval, the function has a value of 1. As x becomes extremely large, the fraction can be simplified to y= x/x = 1. This is because the +2 and -1 don't actually make much difference at high values of x. a billion + one is almost a billion. A trillion + 1 is even more of a trillion. So as you approach infinity, the function approaches x/x and the value becomes 1
To OP, here is a bit of a nifty trick. To find what a function approaches at infinity (or when the value of the top and bottom are both zero, not possible in this case), differentiate the top and bottom, and plug in your value. Here for instance, the derivative of x+1 is 1, the derivative of x-2 is also 1, so if you plug in the value of x ( which is either infinity, or the value of x that generates a 0/0) and you will get your limit. More about it here, http://mathworld.wolfram.com/LHospitalsRule.html
It doesn't tend to Zero. That's wrong. At the beginning of the interval, the function has a value of 1. As x becomes extremely large, the fraction can be simplified to y= x/x = 1. This is because the +2 and -1 don't actually make much difference at high values of x. a billion + one is almost a billion. A trillion + 1 is even more of a trillion. So as you approach infinity, the function approaches x/x and the value becomes 1
To OP, here is a bit of a nifty trick. To find what a function approaches at infinity (or when the value of the top and bottom are both zero, not possible in this case), differentiate the top and bottom, and plug in your value. Here for instance, the derivative of x+1 is 1, the derivative of x-2 is also 1, so if you plug in the value of x ( which is either infinity, or the value of x that generates a 0/0) and you will get your limit. More about it here, http://mathworld.wolfram.com/LHospitalsRule.html