functions help! (1 Viewer)

[Danger]

find f(x) for all x in the following, given that:

a) f(a + 2) = a + 5

b) f(x - 1) = x^2 - 1

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

d) f(t) = at^2 + b

e) f(2) = 5

f) f(3) = 20

i have the answers to these but i want to know HOW to do them please!

 

[tommykins]

Replace all x with all the values inside the brackets.

a) f(a+2) = (a+2) + 5.
b) f(x-1) = (x-1)^2 - 1
c) f(2a) = 4(2a)^2 - 4(2a) + 1
d) f(t) = (t)t^2 + b

for e and f, i'm kind of confused as opposed what to do, as there are no pronumerals.

 

[Danger]

well, answers are

a) f(x) = x + 3
b) f(x) = x^2 + 2x
etc.

so i dont think you just sub in x... its the opposite. anyone know how?

 

[pLuvia]

1) Since f(a+2)=a+5, then that means they subbed in x=a+2 already to obtain that a+5
So that means
f(x)=x+c, where c is some constant
f(a+2)=a+2+c=a+5
c=3
f(x)=x+3

2)
Same reasoning as before
f(x)=x²+c, where c is a constant
f(x-1)=(x-1)²+c=x²-1
x²-2x+1+c=x²-1
c=2x-2
f(x)=x²+2x-2

etc, its the same for the rest

For the last two
treat f(x)=x+c then sub in x=2 and 3 and make f(2)=5 and f(3)=20

 

[Danger]

can you do c) and d) for me aswell?

for the last two i got
f(x)=x+3 and f(x)=x+17

 

[pLuvia]

Those are quite obvious you'll probably be kicking yourself if I show you the solutions lol

Those two are even easier than the first two

Spoiler

c) The only places with the variable "a" are the first and second terms and since they let x=2a, x²=4a², hence f(x)=x²-2x+1

Similarly for d
d) f(x)=ax²+b

And the answer you got are correct

 

[Danger]

sigh.. im still not sure how to do them.. can you go through both of them step by step please?

ok i get it now, thank you so much!

 

[Mattamz]

for e) and f), this is how i think they are done.

e) f(2) = 5 = 2*2.5
therefore f(x) = 2.5x

f) f(3) = 20 = 3*(20/3)
therefore f(x) = 20x/3

 

[Danger]

the results i got for e) and d) is f(x)=x+3 and f(x)=x+17 respectively, thanks anyway