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C3 Maths Question on Functions

Hello,
I am having some difficulty completing the following question in Maths, which relates to functions.



I would really appreciate any help with this question!

Regards,
Aaron.
Original post by Azza.G
Hello,
I am having some difficulty completing the following question in Maths, which relates to functions.



I would really appreciate any help with this question!

Regards,
Aaron.


Which parts have you tried / what is your thought process? :h:
Reply 2
Original post by SeanFM
Which parts have you tried / what is your thought process? :h:



This is my working out so far:

Original post by Azza.G
This is my working out so far:



That is very nice/neat writing :h:

I'll have a look at b, c and d but for part a, think about whether y can really take any value geq 0. For example, when is y = 30?

Also for part b, it's good that you've found the equation for the 'second half' of f(x) but what about the first half? This is important when you're trying to find ff(0). You're better off not finding a formula for ff(x) because it doesn't work like how you've done it unless you carefully define the ranges of x.

Think first about what f(0) is then work out what f(f(0) is.
(edited 7 years ago)
For 4a, not only is the range of f greater than or equal to 0, but it's also less than or equal to 10, since the maximum it ever is within the domain is 10.
For 4b, you only worked out the equation of the line between x = 2 and x =6, the equation is different between x = -2 and x = 2.
Reply 6
Original post by SeanFM
That is very nice/neat writing :h:


Thank you very much! :smile:

Original post by BobBobson
For 4a, not only is the range of f greater than or equal to 0, but it's also less than or equal to 10, since the maximum it ever is within the domain is 10.


Thank you for your input. I have adjusted my range, and you can see it in my working out below.

Original post by SeanFM
That is very nice/neat writing :h:

I'll have a look at b, c and d but for part a, think about whether y can really take any value geq 0. For example, when is y = 30?

Also for part b, it's good that you've found the equation for the 'second half' of f(x) but what about the first half? This is important when you're trying to find ff(0). You're better off not finding a formula for ff(x) because it doesn't work like how you've done it unless you carefully define the ranges of x.

Think first about what f(0) is then work out what f(f(0) is.


I have worked out the equations of both of the lines, but I am not sure how I find f(0). Therefore, I just let y=0 and, and I assume that is f(0)? However, what exactly is f(x), because which equation does it take? I am confused on this part, and do not know what to do next.

Original post by Azza.G
Thank you very much! :smile:



Thank you for your input. I have adjusted my range, and you can see it in my working out below.



I have worked out the equations of both of the lines, but I am not sure how I find f(0). Therefore, I just let y=0 and, and I assume that is f(0)? However, what exactly is f(x), because which equation does it take? I am confused on this part, and do not know what to do next.



Look at your graph given, where is 0, and hence f(0)?
Reply 8
Original post by SeanFM
Look at your graph given, where is 0, and hence f(0)?


When x=0, y=5, and hence then f(0)=5? Is that correct?
Original post by Azza.G
Thank you very much! :smile:



Thank you for your input. I have adjusted my range, and you can see it in my working out below.



I have worked out the equations of both of the lines, but I am not sure how I find f(0). Therefore, I just let y=0 and, and I assume that is f(0)? However, what exactly is f(x), because which equation does it take? I am confused on this part, and do not know what to do next.



You have to let x be 0 and find y in one of the equations. You're doing it the other way round. Doing this will get you different answer for the two different equations, you have to use the appropriate equation. Once you found this, you need to do the same thing with your answer letting x = your answer.
Original post by Azza.G
When x=0, y=5, and hence then f(0)=5? Is that correct?


Correct :h: so ff(0) = ...
Reply 11
Original post by SeanFM
Correct :h: so ff(0) = ...


So, would that mean then ff(0)=5 as well?

Because, f(x)-5/2x+5
So, ff(x) = 25/4 x^2 - 25/2 x +5?
And, when x=0 is subbed in the answer is 5?

Therefore, the original answer of f(0)=5 is derived from the y-intercept. Is this correct?
(edited 7 years ago)
Original post by Azza.G
So, would that mean then ff(0)=5 as well?


Not quite, read ff(0) as f(f(0)). You now know what f(0) is so substitute that into f(f(0)) and then evaluate the next part.
Reply 13
Original post by SeanFM
Not quite, read ff(0) as f(f(0)). You now know what f(0) is so substitute that into f(f(0)) and then evaluate the next part.


Would the answer then be -7.5?
Original post by Azza.G
So, would that mean then ff(0)=5 as well?

Because, f(x)-5/2x+5
So, ff(x) = 25/4 x^2 - 25/2 x +5?
And, when x=0 is subbed in the answer is 5?

Therefore, the original answer of f(0)=5 is derived from the y-intercept. Is this correct?


It is dangerous and unnecessary to find a formula for ff(x).

This is because for different values of x, how you calculate ff(x) will be different. Eg if x = 6 then f(6) = 4 and ff(6) = 2. It is all on the second line.

But when x = 0 f(0) = 5 so it moves from the first line to the second line (to find f(5), you look at the second line).. hence there are multiple parts of f(f(x)) that need to be broken down by range - there isn't a single function that defines ff(x).

So you find f(5) which isn't -7.5

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