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# How do you find the RANGE? C3 watch

1. Something that really bothers me...I've grasped all the hard concepts yet I cant quite get this.... Is there some sort of a direct formula..

Example

Find the range of f(x)= ln(4-2x) x<2 and allreal numbers( X E R)

Thanks
2. Draw the graph. See what values Y can take.
3. I really do agree with you. Range and domain are beyond my understanding. Some graphs are too complicated. Thats one mark I might just let fly tommorrow.
4. The range is what y can be. A normal ln graph can't be less than 0, but since we have -2x, it can't be less than 4. Do you see why?
5. (Original post by face05)
Example

Find the range of f(x)= ln(4-2x) x<2 and allreal numbers( X E R)

Thanks
I still get the range as all real numbers so (X E R). But I could be wrong.
6. Well ln 0 doesn't exist. And you can't have ln of any negative number. Therefore you have to find all values of x where (4-2X) is greater than 0, so you would say X>2.
7. (Original post by mirah12)
I still get the range as all real numbers so (X E R). But I could be wrong.
What it tells you in the question is that the values of x are all real numbers greater than 2. The range is what y can be, the domain is what x can be - which is what the question tells you.
8. (Original post by Jimlon0)
Well ln 0 doesn't exist. And you can't have ln of any negative number. Therefore you have to find all values of x where (4-2X) is greater than 0, so you would say X>2.
that's the domain though, right?
The range for this would be all real numbers, right?
9. The easiest way is to put in some numbers in and make a judgement.
So for example, if the equation was, y=1/(X+1) , and the domain was X>3, Given X+1 will therefore always be positive so if you put it 3, then try 4, you'll see that it will get smaller. So the range would be y<1/4.
If you just look at the equation and just think about what numbers can go in, and therefore what answers you will get out.
10. (Original post by unamed)
that's the domain though, right?
The range for this would be all real numbers, right?
Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
11. (Original post by Jimlon0)
Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
right. Thank you!

I just properly did this, and the poster below me is right. The range would be greater than or equal to for e^x, not lnx
12. This is definitely a mark I'll let fly in the exam.

(Original post by Jimlon0)
Oh yes sorry, I misread the top part. The range would be greater than or equal to 0, because logs don't result in negative numbers.
Assuming that I drew it right...if I draw the graph it passes through the y-axis at ln4 and the x-axis at 1.5 with the asymptote at 2.

The graph then does continue down below 0...

If x=1.9, then f(x)= ln(4-2(1.9)) = - 1.7 (a negative)
13. Range is an element of reals (X E R) for a ln function. In the graph you have given , as x -> 2 , y -> - infinity. and as x -> - infinity , y -> infinity. Therefore y can take any value between negative infinity and infinity, hence all real numbers. Hope this makes sense, and I hope I am not wrong.... else I am in trouble tommorow...
14. (Original post by paper-wings)
Range is an element of reals (X E R) for a ln function. In the graph you have given , as x -> 2 , y -> - infinity. and as x -> - infinity , y -> infinity. Therefore y can take any value between negative infinity and infinity, hence all real numbers. Hope this makes sense, and I hope I am not wrong.... else I am in trouble tommorow...
Yeah that made sense. Major thanks!
15. IMO the easiest way to find it is just to imagine the graph. If it's a transformation of...say...x^2 then I imagine an X^2 graph and shift it along in my head as per the equation. Same with e^x, Ln(X), etc. I know this only works for some people and some equations but...at the end of the day it's usually a 1 mark question. Not worth getting stressed out over.

By the way good luck everyone sitting it tomorrow.
16. I thought the answer to this was f(x) < ln4 ?

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