Fractional Function

By $\href{http://www.artofproblemsolving.com/Wiki/index.php/Jensen's_Inequality}{Jensen's}$ $\text{inequality}$ . a function is either $\text{convex, concave or linear}$ in a specific interval, and depending upon these criteria the inequalities follow:

$f(a_{1}x_1+a_{2}x_2+.....+a_{n}x_n)≤a_{1}f(x_1)+a_{2}f(x_2)+.....+a_{n}f(x_n)$ ....for $f(x)$ being $\text{convex}$

$f(a_{1}x_1+a_{2}x_2+.....+a_{n}x_n)≥a_{1}f(x_1)+a_{2}f(x_2)+.....+a_{n}f(x_n)$ ....for $f(x)$ being $\text{concave}$

$f(a_{1}x_1+a_{2}x_2+.....+a_{n}x_n)=a_{1}f(x_1)+a_{2}f(x_2)+.....+a_{n}f(x_n)$ ....for $f(x)$ being $\text{linear}$

where $\large{a_i \text{ are constants and } \sum\limits_{i=1}^{n}{a_i}=1}$

In this case, we have the third criteria. i.e. that of a $\text{linear}$ function with $n=2$ and $a_1=a_2=\frac{1}{2}$

So, we consider $\large{f(x)=mx+c}$

$\therefore f'(0)=m=-1$

and $f(0)=c=1$

so, our required function:

$f(x)=1-x$

so, $\large{f(2)=\boxed{-1}}$

As f( $\frac{x+y}{2}$ )= $\frac{f(x)+f(y)}{2}$ replacing $x$ by $2x$ and $y$ by 0 we get ....f(x)= $\frac{f(2x)+f(0)}{2}$ or. $f(2x)+f(0)$ = $2f(x)$ or $f(2x)-2f(x)=-f(0)$ .......(equation1).

Now $f'(x)$ = $lim_{h-0}$ $\frac{f(x+h)-f(x)}{2}$ = $lim_{h-0}$ $\frac{f(2x+2h)}{2}$ -f(x) = $lim_{h-0}$ { [f(2x)+f(2h)]/2 -f(x)}/h =limit h tends to 0 { f(2x) + f(2h)-2f(x)}/2h.. =limit h tends to 0 [ f(2h)-f(0)]/2h ......(from equation 1 ) =f'(0)= -1 = -1v x € ( belongs to) R....( given) Integrating we get f (x)= -x +c Putting x=0 we get f(0)= 0+ c = 1 ...(given). So we conclude that c=1 then f(x)= 1- x so f(2)= 1- 2 = -1 hence the answer is -1

By jensen's inequality, f (x) is neither convex or concave I.e. its linear. Considering y=f(x) in cartesian plane , we get any linear equation y=mx+c suffices.so, now procedding further,c=1 and m=-1 hence work is finished.

first $f(\dfrac{x+-x}{2})=\dfrac{f(x)+f(-x)}{2}$ $f(x)+f(-x)=2$ we note that for any x this is true, and hence all term which are power of x cancel each other. but even power of x and -x are same. so, all powers of x are odd.or in other words $f(x)=a_1x^{2n-1}+a_2x^{2n-3}+....+a_{n}x+1$ now $f'(\dfrac{x+-x}{2})=\dfrac{f'(x)+f'(-x)}{2}$ $f'(x)+f'(-x)=-2$ from the above, we see that, $f'(x)=a_1 (2n-1)x^{2n-2}+a_2(2n-3)x^{2n-4}+....+a_n$ by observation, $f'(x)=f'(-x)$ . solving the equations $\begin{cases} f'(x)-f'(-x)=0\\ f'(x)+f'(-x)=-2\\ \end{cases}$ we add these 2 to get $f'(x)=-1\rightarrow a_n=-1$ since $f'(x)$ is a constant polynomial, $f(x)$ must be of the 1st degree.we see that $f(x)=a_nx+1=1-x$ and $f(2)=1-2=\boxed{-1}=f'(2)$

algebra method(put x=2x,y=0)....

2f(x)=f(0)+f(2x)

f(2x)-f(x)=f(x)-f(0x)

hence f(0x),f(1x),f(2x) in AP

therefore

f(x) = A+Bx..now just solve

My method used partial differentiation such that x is treated as a constant so f(x) is treated as a constant (Note: f( $\frac{x+y}{2}$ ) is NOT treated as a constant ). Anyway, differentiating in respect to x: $f'(\frac{x+y}{2}\ ) = \frac{ f'(x)}{2} \Rightarrow\ let \ x=0 \Rightarrow\ f'( \frac{y}{2} ) = \frac{f'(0)}{2} = - \frac{1}{2}\$ Now integrating produces: $f( \frac{y}{2} ) = - \frac{1}{2}\ y + K \Rightarrow\ let \ y=0 \Rightarrow\ k= f(0) = 1$ $\therefore\ f(y) = 1-y \ so \boxed{ f(2) = -1 }$

You can just guess $-x+1$ and it works...