Composite function of a piecewise function

libragirl79
Messages
30
Reaction score
0

Homework Statement



Given that I have a doubling function :
f(x)=2x (for 0≤x<0.5) and 2x-1 (for 0.5≤ x<1)

Homework Equations


What is f(f(x))?


The Attempt at a Solution


f(f(x))=4x for the first one and 4x-3 for the second part but not sure what to do about the domain constraints...

Thanks!
 
Physics news on Phys.org
libragirl79 said:

Homework Statement



Given that I have a doubling function :
f(x)=2x (for 0≤x<0.5) and 2x-1 (for 0.5≤ x<1)

Homework Equations


What is f(f(x))?


The Attempt at a Solution


f(f(x))=4x for the first one and 4x-3 for the second part but not sure what to do about the domain constraints...

Thanks!
It's not that simple.

Yes, f(f(x)) = 4x over part of the domain of the composite function and 4x-3 over some other portion, but those are not the only two pieces of f(f(x)).


For what values of x is 0 ≤ 2x < 0.5 ?

For what values of x is 0.5 ≤ 2x < 1 ?

etc.
 
right, that's exactly my issue, since i don't have the fcns for 1/4 ≤ x < 1/2
and 1/2 ≤ x < 3/4 ... is there a certain method for doing this?
 
libragirl79 said:
right, that's exactly my issue, since i don't have the fcns for 1/4 ≤ x < 1/2
and 1/2 ≤ x < 3/4 ... is there a certain method for doing this?
You understand that "1/4\le x&lt; 1/2" is part of the interval 0\le x&lt; 1/2 don't you?

If 0\le x&lt; 1/4, f(x)= 2x which is less than 1/2 so ff(x)= f(2x)= 2(2x)= 4x.
If 1/4\le x&lt; 1/2 then x is still between 0 and 1/2 so f(x)= 2x but f(x) is now between 1/2 and 1 so ff(x)= f(2x)= 2(2x)- 1= 4x- 1.

Do similarly for 1/2\le x&lt; 3/4. Now x is between 1/2 and 1 so f(x)= 2x- 1 which is between 0 and 1/2.

If x is between 3/4 and 1, f(x)= 2x- 1 is between 1/2 and 1
 
I understand the breakdown of the x domains, but how do you know where f(x) falls?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top