Help with Functions (2 Viewers)

[noobonastick]

Hi can someone help me with these problems, yes they are probably basic... but i was away when the teacher explained (or wasnt listening ) . So yah, help me

I have copied the question exactly; (ambiguous question imo, dont really know what its trying to ask)

Find f(x) for all x in each of the following, given that

1) f(x-1) = x^2 -1
2) f(2a) = 4a^2 - 4a + 1
3) f(t) = at^2 + b, f(2) = 5, f(3) = 20

the answers to 1,2,3 are respectively

f(x) = x + 2x
f(x) = x^2 - 2x + 1
f(x) = 3x^2 - 7

 

[Iruka]

I'll show you how to do the first one. I think you'll see how to do the second one yourself then.

We are given f(x-1) = x^2 -1 (1)

Let y=x-1. Then x=y+1, and subsituting into (1), we get
f(y) = (y+1)^2 -1
=y^2 +2y +1 -1
= y^2 +2y

But y is just a "dummy variable," so we can really call it anything we like. So let's rename it x again, and then you can see that

f(x) = x^2 +2x.

Part 3 is just solving the two simultaneous equations for a and b that you obtain when you substitute the given values of t into the function.

 

[noobonastick]

so for 2 it would be ;

f(2a) = 4a^2 -4a + 1 (1)

let x = 2a, then a = x/2

sub a= x/2 into (1)

which then equals x^2 - 2x +1 (after working out)

just checking if i did it right lol.

and Part 3 is simple

 

[lyounamu]

so for 2 it would be ;

f(2a) = 4a^2 -4a + 1 (1)

let x = 2a, then a = x/2

sub a= x/2 into (1)

which then equals x^2 - 2x +1 (after working out)

just checking if i did it right lol.

and Part 3 is simple

Yes, you are right. But don't forget to check that,
f(2a) = (2a)^2 - 2(2a) + 1
= 4a^2 - 4a + 1

 

[noobonastick]

yea thx for the help