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# C3 functions watch

1. h(x) is defined by

h(x) = (2x +7)/(5x-2)

show that hh(x) = x

I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
2. (Original post by ihatePE)
h(x) is defined by

h(x) = (2x +7)/(5x-2)

show that hh(x) = x

I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
are you allowed to say let x = 1 then plug 1 in (which gives you 3). then plug 3 back in and you get 1
3. Try rewriting ur numerator and denominator as one fraction for each, with the numerator as (5x-2) for both.

It should cancel nicely from there.

(Original post by ihatePE)
h(x) is defined by

h(x) = (2x +7)/(5x-2)

show that hh(x) = x

I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
If u still dont know how to proceed let me know.👊🏾
4. (Original post by Shaanv)
Try rewriting ur numerator and denominator as one fraction for each, with the numerator as (5x-2) for both.

It should cancel nicely from there.

If u still dont know how to proceed let me know.👊🏾
when separating I got (2x)/(5x-2). and 7/(5x-2)
so for (2x)/(5x-2)
I got 4x+14 when simplified (?) I think I done this wrong as well

for 7/(5x-2)
I got (35x-14)/(10x+35) which is where I'm stuck again because I can't see any common factors to simplify this fraction
5. Its really hard for me to explain with words, and im no good at latex so i hope this helps.

6. (Original post by ihatePE)
when separating I got (2x)/(5x-2). and 7/(5x-2)
so for (2x)/(5x-2)
I got 4x+14 when simplified (?) I think I done this wrong as well

for 7/(5x-2)
I got (35x-14)/(10x+35) which is where I'm stuck again because I can't see any common factors to simplify this fraction
Sorry forgot to quote u. ^^^^
7. (Original post by Shaanv)
Its really hard for me to explain with words, and im no good at latex so i hope this helps.

sorry I can't see the pic
8. top of the fraction: 2(2x+7)/(5x-2) + 7 which simplifies to (4x+14)/(5x-2) + 7 change 7 into 7(5x-2)/(5x-2)
so you get (4x+14 +35x-14)/5x-2 = 39x/(5x-2)

bottom of fraction:
5(2x+7)/(5x-2) -2 which simplifies to (10x+35)/5x-2) -2 change -2 into -2(5x-2)/(5x-2)
so you get (10x+35-10x+4)/(5x-2) = 39/(5x-2)

then cancel the 5x-2 because its on the top and bottom so you get 39x/39 which = x

9. (Original post by ihatePE)
sorry I can't see the pic
10. There’s an easier way. Consider what hh(x) means. It means that if you start with x, run it through function h, then put the result of that through h again you get back to x. That can only happen if h is the same as [inverse h]. So all you need to do is show that [inverse h](x) = h(x) and you’re done.
11. (Original post by Cor395)
are you allowed to say let x = 1 then plug 1 in (which gives you 3). then plug 3 back in and you get 1
that is a useful check, but it is not a proof. you need to show that it is true for all values of x in the domain.
12. I mean it's pretty easy to multiply the numerators in your head to get

. After that you just multiply numerator and denominator by 5x - 2.

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