A composition of function problem

[Rron]

Homework Statement

We have f(f(x))=4x-15 , what is f(2)?

Homework Equations

Don't know.

The Attempt at a Solution

Don't know how to start actually!

 

[Curious3141]

Homework Statement

We have f(f(x))=4x-15 , what is f(2)?

Homework Equations

Don't know.

The Attempt at a Solution

Don't know how to start actually!

Let f(x) = ax + b, and see where you go from there.

There are 2 distinct answers here.

 

[Rron]

Curios3141 thanks but couldn't get anything.
Tried a lot.

 

[HallsofIvy]

Then show us what you tried! If f(x)= ax+ b, what is f(f(x))?

 

[Rron]

Then show us what you tried! If f(x)= ax+ b, what is f(f(x))?

this is what i tried:
if f(x)=ax+b then f(2)=2a+b
I substituted f(x) in f(f(x)) with ax+b so it becomes f(ax+b)=4x-15
Now ax+b=2 so from this x=2-b\a so if u substitute x in the equation above f(2)=4(2-b\a)-15
then I equalized 4(2-b\a)-15=2a+b, but from this u can't get anything.

 

[SammyS]

this is what i tried:
if f(x)=ax+b then f(2)=2a+b
I substituted f(x) in f(f(x)) with ax+b so it becomes f(ax+b)=4x-15
Now ax+b=2 so from this x=2-b\a so if u substitute x in the equation above f(2)=4(2-b\a)-15
then I equalized 4(2-b\a)-15=2a+b, but from this u can't get anything.

Hello Rron. Welcome to PF !

What is f(f(x)), using only the assumption that f(x) = ax+b ?

 

[Curious3141]

Curios3141 thanks but couldn't get anything.
Tried a lot.

As others have already stated, derive an expression for f(f(x)) in terms of a,b and x. Then set this equal to 4x-15 and see what values a and b can take.

 

[spaghetti3451]

I hope no one would mind me writing down the first few lines of the solution to guide our friend here.

Assume that f(x) = ax+b. We need to find f(2), which, from the assumption, equals 2a+b. This means we need to know the values of a and b to find f(2).

To find a and b, substitute f(x) = ax+b in the equation f(f(x)) = 4x-15. You have a[f(x)]+b = 4x-15 i.e. ...

 

[SammyS]

I suppose we should wait for OP to show up in this thread before anyone else posts.

If he doesn't post in a week or two it might be a good idea for one of us to finish this up.

What does the management think about that?

 

[Rron]

Sorry but still nothing. Maybe it is because I learned these a long time ago. It actually has been 3 years since I last solved a function problem like this. So can you please show me the way that you solved this problem in order to save time struggling with the problem and then at the end getting nothing.
Thanks.

 

[Curious3141]

Sorry but still nothing. Maybe it is because I learned these a long time ago. It actually has been 3 years since I last solved a function problem like this. So can you please show me the way that you solved this problem in order to save time struggling with the problem and then at the end getting nothing.
Thanks.

Ah, but if we just present the solution, will you learn anything?

Let's take it step-by-step. It's just algebra.

Start with f(x) = ax+b

Then f(f(x)) = a(ax+b) + b = ?

We'll take it from there after you expand the bracket and rearrange terms to get that expression.

 

[Rron]

Ah, but if we just present the solution, will you learn anything?

Let's take it step-by-step. It's just algebra.

Start with f(x) = ax+b

Then f(f(x)) = a(ax+b) + b = ?

We'll take it from there after you expand the bracket and rearrange terms to get that expression.

That's a piece of cake man. a^2x+ab+b

 

[Curious3141]

That's a piece of cake man. a^2x+ab+b

Great. Now compare that with what the question gave: f(f(x)) = 4x-15

So,

a²x + (ab + b) = 4x - 15

For that to be true in general, the coefficients of each term have to be equal. So you can state:

a² = 4 ---equation 1

ab + b = -15 ---equation 2

Can you solve that system of simultaneous equations?

 

[Rron]

Curios3141 thank you so much finally solved it.
The answer is -1.

 

[Curious3141]

Curios3141 thank you so much finally solved it.
The answer is -1.

It's not the only answer though. Remember a = ±2. b can similarly take different values in each case. You get two perfectly valid forms for f(x). f(2) can take different values depending on which form.

 

[Rron]

Yeah I know that.-1 and 9 but forgot to tell you that I got some choices:
A)-1
B)-2
C)-3
D)-4

 

[Curious3141]

Yeah I know that.-1 and 9 but forgot to tell you that I got some choices:
A)-1
B)-2
C)-3
D)-4

9?

The other possible f(2) should be 11.

f(x) can be 2x - 5 → f(2) = -1

f(x) can be -2x + 15 → f(2) = 11

But here, you go with choice A of course.