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# C3 range, domain question watch

1. Hey guys.

I always get stuck on questions involving the range, domain, inverse functions etc. I am doing a C3 past paper(June 2008) and I am stuck at Q4( below). Please tell me how to find the range and the domain i.e. parts b) and c).

Thank you in advance! (+ve rep for the most helpful person )
2. Don't see the link?
3. (Original post by boromir9111)
Don't see the link?
oh yes, sorry, silly me!

4. The function f is defined by : f(x)= 2(x-1)/(x^2-2x-3) - 1/(x-3), x>3

(a) Show that f(x)= 1/(x+1), x>3 (4) I have done this.

(b) Find the range of f. (2)

I have tried to sub in x=3 to get f(x)=1/4 but not sure how that works.
I also have sketched the graph and it is clear that f(x)>0, but don't know where to go from there.

(c) Find f –1 (x). State the domain of this inverse function.(3)

I know that the inverse function is f-1(x)=(1-x)/x and I know that the range of the function is the domain of the inverse function , so I only need help with part b)
4. (Original post by Gabriela)
oh yes, sorry, silly me!

4. The function f is defined by : f(x)= 2(x-1)/(x^2-2x-3) - 1/(x-3), x>3

(a) Show that f(x)= 1/(x+1), x>3 (4) I have done this.

(b) Find the range of f. (2)

I have tried to sub in x=3 to get f(x)=1/4 but not sure how that works.
I also have sketched the graph and it is clear that f(x)>0, but don't know where to go from there.

(c) Find f –1 (x). State the domain of this inverse function.(3)
(b) the range is simply the values you can get out when values from the domain are put in.

now your domain is x>3 , you have identified f(3) = 1/4 ... either from your graph or by putting in other valid values of x you should see that all subsequent values are
(a) positive and
(b) less than 1/4

so range is 0 < f(x) < 0.25

for part (c)

write y= 1/(x+1)

now swap x's and y's

x=1/(y+1)

now make y the subject and finally re-write replacing y= with notation for inverse =
5. (Original post by Gabriela)
oh yes, sorry, silly me!

4. The function f is defined by : f(x)= 2(x-1)/(x^2-2x-3) - 1/(x-3), x>3

(a) Show that f(x)= 1/(x+1), x>3 (4) I have done this.

(b) Find the range of f. (2)

I have tried to sub in x=3 to get f(x)=1/4 but not sure how that works.
I also have sketched the graph and it is clear that f(x)>0, but don't know where to go from there.

(c) Find f –1 (x). State the domain of this inverse function.(3)
For (b):

Notice that as x increases, f(x) gets smaller. As you correctly noticed however, it never reaches f(x) = 0.
If the range is the set of values f(x) can take, how could you express that?

For (c):

Do you know how to find the inverse of a function, or is purely the domain part you're struggling with?
6. (Original post by H.C. Chinaski)
(b) the range is simply the values you can get out when values from the domain are put in.

now your domain is x>3 , you have identified f(3) = 1/4 ... either from your graph or by putting in other valid values of x you should see that all subsequent values are
(a) positive and
(b) less than 1/4

so range is 0 < f(x) < 0.25

for part (c)

write y= 1/(x+1)

now swap x's and y's

x=1/(y+1)

now make y the subject and finally re-write replacing y= with notation for inverse =
Thank you
7. (Original post by EEngWillow)
For (b):

Notice that as x increases, f(x) gets smaller. As you correctly noticed however, it never reaches f(x) = 0.
If the range is the set of values f(x) can take, how could you express that?

For (c):

Do you know how to find the inverse of a function, or is purely the domain part you're struggling with?
Thank you, I have figured it out now, but I appreciate your help and will give you +ve rep too

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