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Maths help - ranges of functions

It would be wonderful if anyone could help me with this:

Functions g and are defined by g(x)=√x and h(x)=(2x-3)/(x+1).


Find the range of h.


Find the domain and range of goh.

For this I've got as far as goh=√(2x-3)/(x+1)

But not sure on how to find the ranges or domains :frown:
Reply 1
Avatar for EconFan_73
EconFan_73
3
When finding the range of a function like that, you've got to rearrange so that you get a known constant added to a fraction with the unknown in.

In this case, h(x) can be rearranged to (2(x+1)-5)/(x+1) = 2 - 5/(x+1)

As 5/(x+1) cannot equal zero, but can equal anything else, the range of h(x) is that y can be any real number, apart from 2.

As the range of h(x) is then input into g(x), the domain of gh(x) is the same as the range for h(x). (The previous output is the new input)

Now that you know the domain for gh(x), I expect that you'll be able to work out the range! :smile:

Hope this helps!
(edited 10 years ago)
Original post by EconFan_73
When finding the range of a function like that, you've got to rearrange so that you get a known constant added to a fraction with the unknown in.

In this case, h(x) can be rearranged to (2(x+1)-5)/(x+1) = 2 - 5/(x+1)

As 5/(x+1) cannot equal zero, but can equal anything else, the range of h(x) is that y can be any real number, apart from 2.

As the range of h(x) is then input into g(x), the domain of gh(x) is the same as the range for h(x). (The previous output is the new input)

Now that you know the domain for gh(x), I expect that you'll be able to work out the range! :smile:

Hope this helps!


How did you get this (bolded bit)??

And no, on all the other questions I have been able to find the domain with ease, but had severe difficulty finding the range..
Reply 3
Avatar for EconFan_73
EconFan_73
3
Original post by ILovePancakes
How did you get this (bolded bit)??

And no, on all the other questions I have been able to find the domain with ease, but had severe difficulty finding the range..


Basically you rewrite your numerator in terms on your denominator.

So (2x-3) can be rewritten as 2(x+1) - 5. You then use this as the numerator.

So you have (2(x+1)-5)/(x+1)

This can be split up into (2(x+1))/(x+1) - 5/(x+1)

This can then be simplified to 2 - 5/(x+1)

It might help if you rewrite what I've put out on paper as fractions with numerators and denominators, so you can see how I've cancelled etc :smile:
Original post by EconFan_73
Basically you rewrite your numerator in terms on your denominator.

So (2x-3) can be rewritten as 2(x+1) - 5. You then use this as the numerator.

So you have (2(x+1)-5)/(x+1)

This can be split up into (2(x+1))/(x+1) - 5/(x+1)

This can then be simplified to 2 - 5/(x+1)

It might help if you rewrite what I've put out on paper as fractions with numerators and denominators, so you can see how I've cancelled etc :smile:


Thank you! You've been really helpful :biggrin:
Reply 5
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EconFan_73
3
Glad to help!:P

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