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# C3-function question watch

1. The function f has a domain x ≥ 0 and is defined by

f(x) = 8 / x+2

Find an expression for f-1(x) and write the domain of f-1.

y= 8 / x+2

y (x+2) = 8

x+2= 8/y

x= 8/y - 2

therefore f-1(x) = 8/x - 2

domain x ≥4

Would that be a right answer? And how do I find domain in this case, im quite unsure.
2. y= 8 / x+2

yx+2y = 8

yx = 8-2y
x = 8-2y/y

f-1(x) = 8-2x/x

Domain for f-1(x) is the range for f(x)

Draw a sketch to find the range of f(x).

As above, it is the range of f(x).

You're almost right. The 4 is right anyway, but you need to check again. What happens as you input values bigger than 2? So what does that make the domain?
4. (Original post by FZka)
y= 8 / x+2

yx+2y = 8

yx = 8-2y
x = 8-2y/y

f-1(x) = 8-2x/x

Domain for f-1(x) is the range for f(x)

Draw a sketch to find the range of f(x).

is the range for f(x) ≥ 4 ???
5. (Original post by AnonyMatt)

As above, it is the range of f(x).

You're almost right. The 4 is right anyway, but you need to check again. What happens as you input values bigger than 2? So what does that make the domain?
I am too confused now...could you just please tell me the method how to get domain please?
6. Substitution doesn't always work. Try to make a sketch.

The horizontal asymptote would be the line y=0 ( the x-axis)
The vertical asymptote would be x= -2.

i.e. the range for values of x>0 is f(x)>0
7. (Original post by Janka3112)
I am too confused now...could you just please tell me the method how to get domain please?
I did when I wrote "What happens to f(x) if you put in values bigger than 0?"*

*Realised in my other post I said 2. Oops. Doesn't matter, same principle anyway.

Right, your post says that the domain for f(x) is x>=0, right?

EDIT: Deleted. Mega mega facepalm.

(Original post by FZka)
Substitution doesn't always work. Try to make a sketch.

The horizontal asymptote would be the line y=0 ( the x-axis)
The vertical asymptote would be x= -2.

i.e. the range for values of x>0 is f(x)>0
Of course. I read the fraction wrong. Faaaaaaaaacepalm.

OP. Put in x = 0.1
Notice it takes a value greater than 4.
Edit: No it doesn't. Why am I saying these things?
I must be so confusing right now. I'm confused myself!
So, I think that f(x) must be <=4 too... Maybe. Yes, that's right.

So the range is 0<f(x)<=4? I think so.

Maybe it would be better to take the other poster's advice and sketch the graph. I hate sketching so I just plug numbers in. I told you the wrong way to do it though. Anyway, as the above poster showed you, f(x) can never be less than 0, nor actually 0. But it is always greater than 0.

I'm leaving now. Making far too many mistakes tonight. Ugh.
8. (Original post by AnonyMatt)
I did when I wrote "What happens to f(x) if you put in values bigger than 0?"*

*Realised in my other post I said 2. Oops. Doesn't matter, same principle anyway.

Right, your post says that the domain for f(x) is x>=0, right?

EDIT: Deleted. Mega mega facepalm.

Of course. I read the fraction wrong. Faaaaaaaaacepalm.

OP. Put in x = 0.1
Notice it takes a value greater than 4.

Maybe it would be better to take the other poster's advice and sketch the graph. I hate sketching so I just plug numbers in. I told you the wrong way to do it though. Anyway, as the above poster showed you, f(x) can never be less than 0, nor actually 0. But it is always greater than 0.

thank you so much, finally some good explanation,which i understood, thanks, u saved me from going mental
9. (Original post by Janka3112)
thank you so much, finally some good explanation,which i understood, thanks, u saved me from going mental
Read my new editted post. Some of it was wrong in the one you quoted.

Just remember:

domain of f(x) = range of f-1(x)
range of f(x) = domain of f-1(x)

Range of f(x) is all the values it can take, given the domain.
Plug in the lower value (in your question, 0) and higher value (infinity) and see what the range is.

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