Cool function

$f(f(x))=1-x$ Replacing $x$ by $f(x)$ , we get $f(f(f(x)))=1-f(x)$ $\Rightarrow f(1-x)=1-f(x)$ putting $x=\frac{1}{2}$ , we get $2f(\frac{1}{2})=\boxed 1$

$f(f(x))=1-x$ $f(f(\frac { 1 }{ 2 } ))=1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 }$

$f(\frac { 1 }{ 2 } )=f(f(f(\frac { 1 }{ 2 } )))=1-f(\frac { 1 }{ 2 } )$ $f(\frac { 1 }{ 2 } )=1-f(\frac { 1 }{ 2 } )\\ 2f(\frac { 1 }{ 2 } )=1$

here x=1/2, so that 2f(1/2)=2 (1-1/2)=2 (2-1/2)=2*1/2=1

Since it seems that this will hold for any function $f$ , a strategy we can do is to pick any function satisfying the given relation. But a particular formula of a function $f$ satisfying the relation is not immediately apparent. Instead, let's say $f$ is any function which has an inverse. If $f(x) = 1/2$ then $x = f^{-1}(1/2)$ , so $f(1/2) = 1 - f^{-1}(1/2)$ . On the other hand, taking $f^{-1}$ on both sides of the given equation, we get $f(x) = f^{-1}(1-x)$ . Plug in $x=1/2$ , we have $f(1/2) = f^{-1} (1/2)$ . Since $f(1/2) = 1 - f^{-1}(1/2)$ and $f(1/2) = f^{-1} (1/2)$ , then $f(1/2)=1/2$ , so $2f(1/2)=1$ .

Let $f(x)=ax+b$

$f(f(x))=a(ax+b)+b$

$f(f(x))=a^2x+b(a+1)=1-x$

$a=i$ and $b=\frac {1}{i+1}$

$f(x)=ix+\frac {1}{i+1}$

$2f(\frac {1}{2})=2(i\times \frac {1}{2}+\frac {1}{i+1})$

$2f(\frac {1}{2})=(i+\frac {2}{i+1})$

$2f(\frac {1}{2})=\frac {i(i+1)+2}{i+1}$

$2f(\frac {1}{2})=\frac {i+1}{i+1}=\boxed{1}$

$f(f(\frac{1}{2})) = 1 - \frac{1}{2} = \frac{1}{2}$ $f(f(x)) = 1 - x \Rightarrow f(f(f(x))) = 1 - f(x) \iff f(x) = 1 - f(f(f(x)))$ $f(\frac{1}{2}) = 1 - f(f(f(\frac{1}{2}))) = 1 - f(\frac{1}{2}) \iff 2 f(\frac{1}{2}) = 1$

f(f(x)) = 1 - x, so f(f(f(x))) = x. With f(f(1/2)) = 1 - 1/2, we have f(1/2) = f(f(f(1/2))) = 1/2. Done.

Rather weird (and wrong) solution.

If we let $f(x) = x$ for some $x$ , we get the equation

$x = 1-x$

$\displaystyle x = \frac{1}{2}$

Therefore, $f(x) = x$ if $\displaystyle x = \frac{1}{2}$

$\therefore \displaystyle 2f(\frac{1}{2}) = 2 \times \frac{1}{2} = \boxed{1}$

Just put x=1/2 in the first equation, ans automatically comes to be 1.

Let f(f(x))=1/2=1-x, so x=1/2. This gives f(1/2)=1/2. Hence, 2f(1/2)=1

This problem was sooooooo easy!!!!! I really did like this problem though so thank you!!!!!!! :-)

If you divide the left and right side of the equation, you get the function half, and the actual quantity half. f(f(x)) -- the function half 1-x -- the quantity Functions that are written using f(x) are just over-formalized ways of indicating what the independent variable of the function is. The overall point is that x needs to be given a value, and its value will affect the value of the quantity. For example, x=1, 1-x, 1-1=0 The function side of the equation would be written like this f(f(1)) But regarding this part just confuses the problem.

So, we are told to find the value at 2f(1/2) Rewriting the original equation, we get 2f(1/2) = 2(1- 1/2) =1

take f(f(x))=f(1/2) then,f(1/2)=1-1/2=1/2 now ,multiply,2 and 1/2 it results 1.

from f(f(x)) = 1 - x, let x = f(x). Then, f(x) = 1 - f(x). Separating f(x) we get 2f(x) = 1. If x = 1/2 2f(1/2) = 1.

f(f(x)) = 1-x

Find 2f(1/2)

1) From observation, f(1/2) which is equivalent to f(x). Since f(x)=f(1/2) hence x=1/2

2) Substitute x = 1/2 then the original function would be f(1/2) = 1-1/2

3) Get the required: 2f(1/2) = 2(1-1/2) = 1

Therefore the answer is 1

since fof(x)=x, we get x=1/2 therofore fof(x)=1

f(f(x)=1-x put x=1/2 f(1/2)=1/2 2f(1/2)=2(1/2)=1

All you had to do was substitute x for 1/2 and multiply by 2

Since f(x) = 1-x thus f(1/2) = 1-1/2=1/2 2f(1/2)= 2x1/2 = 1

2f(1/2)=2(1-1/2)=1

f(f(x)) = f^2(x) f^2(x) = 1 - x than by dividing by x on both sides shows that f^2 = 1 Since the square root of 1 is 1 and the square root of f^2 is f than f = 1 Than 2f(1/2) is 2(1)(1/2) = 1

f(f(x))=f(1-x)
f(1-x)=1-(1-x)=x
2(f(1/2))=2*1/2
=1

when x=1/2 then f(1/2)=1/2 now,f(x=1/2)=1/2 & f=2 so after multiplying f*f(1/2)=1

Write a solution. First solve for x=1/2 f(f(1/2))=1-1/2=1/2 , then multiply by 2, 1/2x2=1

f(1)=2(1-1/2) =2(1/2) =1

f(f(x))=1-x fsquaredx=1-x ----- ---- x x fsquared=1 f=1squared f=1 value of 2f(½) =2x1(½) =2(½) =1

Let f(x) =ax+b ; It is obvious the the function is liner one. Then f(f(x)) = a(ax+b)+b = a^2 x+ab+b Now, a^2 x+ab+b = 1-x ; Making an identity Comparing the co-efficient of x & constant term; a^2 = -1 & ab + b = 1 Solving these equation, a= i & b = (1 - i)/2 ; Taking only positive solution So, f(x) = ix +(1-i)/2 2f(1/2) =2[ i(1/2) + (1-i)/2] = i +(1-i) =

If we are giving the vale x=1/2 , we get f( f(1/2))=1/2 from first equation, and the second equation becomes 2 f(1/2)=2(1-1/2)=2(1/2)=1. Finally answer is 1.

Assume f(x)=y

So f(y)=1-x

Calculate invert function

So assume f(y)=z

z=1-x

x=1-z

So f(x)=1-x

2f(1/2)=1

This is right method?

This equation is a nestid loop. Therefore we now that f(f(x))= f(x) so 2(1- (1/2)) = 1

f（f(x)=1-x f(x)=x f(1/2)=1/2 2*1/2=1