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C3 functions

ff(0) is just f[f(0)]. Find f(0) first, which in this case is 5. Now find f(5) on this same graph.

Spoiler

Reply 2
Original post by meowgoesthedog
ff(0) is just f[f(0)]. Find f(0) first, which in this case is 5. Now find f(5) on this same graph.

Spoiler


i don't understand where the 5 comes from how can one know it's 5?
Original post by bigdonger
i don't understand where the 5 comes from how can one know it's 5?


I see is as the function being made up of two straight lines for 2 separate ranges of x values. So when -2<x<2 f(x) = 2.5x + 5, but when 2<x<6 f(x) = x-2. Since -2<0<2 then it lies on the first line. f(0)=5. f(5) lies on the other line since 2<5<6 so f(5) is 5-2=3.
[QUOTE=bigdonger;66436108]i don't understand where the 5 comes from how can one know it's 5?

You are given points which you can use to construct an equation of the line from X=-2 to X=2. Then just plug in X=0 to get the 5
Reply 5
Original post by RDKGames
You are given points which you can use to construct an equation of the line from X=-2 to X=2. Then just plug in X=0 to get the 5


ok thanks and then what about the next bit?
Original post by bigdonger
ok thanks and then what about the next bit?


Same thing with the right hand side of the function that has the positive gradient.
Reply 7
Original post by RDKGames
Same thing with the right hand side of the function that has the positive gradient.


so i get the equation of the line then set y =0?
i really don't understand what i'm doing here

Edit: sorry i got it now, you set x=5 and i got 3 thanks :smile:
(edited 7 years ago)

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