Composite Function Problem

[turophile]

The problem:

Give an example of a function f for which f[f(x)] = 2x + 4.

Any hints about a type of function that would work here?

 

[galactus]

They are asking you to find a function such that when you enter that function back into itself you get 2x+4.

Here is an example of what I mean.

Take f(x)=2x+4

Plug 2x+4 in for x back into 2x+4

2(2x+4)+4=4x+12

See?. Only here we get 4x+12, not 2x+4.

The trick is to find one that gives you 2x+4 when doing this.

The thing to do is try equating coefficients.

Take f(x)=ax+b. We know that if we sub in x=ax+b into f(x) we have to get 2x+4. Follow so far?.

a(ax+b)+b=2x+4

Let me know how you proceed.

 

[turophile]

I see it now.

a(ax + b) + b = a[sup:9y1us90t]2[/sup:9y1us90t]x + ab + b = 2x + 4

So a = sqrt(2). Then sqrt(2) * b + b = [1 + sqrt(2)] * b = 4, and b = 4/[1 + sqrt(2)].

Checking to make sure it works:

sqrt(2) * [ sqrt(2) * x + 4/(1 + sqrt(2)) ] + 4/[1 + sqrt(2)]

= 2x + [4 * sqrt(2)]/[1 + sqrt(2)] + 4/[1 + sqrt(2)]

= 2x + [4 * sqrt(2) + 4]/[1 + sqrt(2)]

= 2x + 4 * [1 + sqrt(2)]/[1 + sqrt(2)]

= 2x + 4

Many thanks, galactus.

 

[galactus]

Very good.

I am glad to see you got it.

May I ask what class this is?. You have posted some interesting problems. Like the one with the two parabolas and the tangents.

 

[turophile]

I'm reviewing problems from my old college calculus textbook as a refresher for getting back into mathematics. I'm going to try to get back up to speed with calculus so I hopefully won't have to retake the courses that I took back in the 1970's before going on in math. I took horrible class notes back then, so I'm struggling with some of the more advanced end-of-section exercises. When I've gotten stuck, I've posted them here. What a great resource this site is.