# Homework Help: Differentiability and functional equations

1. Jul 1, 2017

### Tanishq Nandan

1. The problem statement, all variables and given/known data
Let f((x+y)/2)= {[f(x)+f(y)]/2} for all real x and y
{f'(x)=first order derivative of f(x)}
f'(0) exists and is equal to -1 and f(0)=1.
Find f(2)
2. Relevant equations
Basic formula for differentiablilty:
f'(x)=limit (h tends to 0+) {[f(x+h)-f(x)]/h}

3. The attempt at a solution
I know that when you have a functional equation along with some info about it's derivative,you need to apply the basic formula of differentiablilty to find f'(x) and evaluate the limits using the given functional equation..and that's precisely what I did..but how do I get to f(2) from this?? Hope you get what I did in the second last step..I used the given functional equations,placed x=2h and y=0,then replaced the value of f(h) obtained from there in the given limit.
Now,as I said,how to get f(2) from here??

2. Jul 1, 2017

### Staff: Mentor

Thread has been moved. Please post questions that involve derivatives in the Calculus & Beyond section.
I don't think your approach using the definition of f'(0) will help you. From the given information, $\frac 1 2 \left(f(1) + f(-1)\right) = 1$, or equivalently
$f(1) + f(-1) = 2$. In addition, $f(1/2) + f(-1/2) = 2, f(2) + f(-2) = 2$, and so on. Numbers that are opposites always have function values that add up to 2. What sort of curves have this property?

3. Jul 1, 2017

### Tanishq Nandan

The function is odd about x=1 ! !
f(2) will correspond to negative of f(0)
Ans (-1) (and the answer is matching)
But,just one slight problem..where did we make use of f'(0)=-1 ? ?

4. Jul 2, 2017

### pasmith

$$f\left(\frac{(1 + x) + (1 - x)}{2}\right) = \frac{f(1 + x) + f(1-x)}{2}$$ gives you $f(1 + x) + f(1-x) = 2f(1)$, which doesn't help you as you don't know what $f(1)$ is.

Your functional equation is actually telling you that $f$ is both concave and convex. What functions have that property?

5. Jul 5, 2017

### Tanishq Nandan

A straight line?
Y=mx+c??