MHB Functional Equation: Solving for f(2012)

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The functional equation \(f(x+y) = f(xy)\) leads to the conclusion that \(f\) is a constant function. Given that \(f\left(-\frac{1}{2}\right) = -\frac{1}{2}\), it follows that \(f(x) = -\frac{1}{2}\) for all \(x\). Consequently, \(f(2012)\) must also equal \(-\frac{1}{2}\). The reasoning confirms that the function does not vary with different inputs. Therefore, \(f(2012) = -\frac{1}{2}\).
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If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $
 
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jacks said:
If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

Hi jacks, :)

Since, \(f(x+y) = f(xy)\) we have,

\[f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}+0\right)=f\left(-\frac{1}{2}\times 0\right)=f(0)\]

Since, \(f\left(-\dfrac{1}{2}\right) = -\dfrac{1}{2}\)

\[f(0)=-\frac{1}{2}\]

Now,

\[f(2012)=f(2012+0)=f(2012\times 0)=f(0)=-\frac{1}{2}\]

Kind Regards,
Sudharaka.
 
jacks said:
If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

Observe:

\[f(x+y)=f(xy) \Rightarrow f(x)=f(0)\]

Hence \( f(x) \) is a constant function, so \(f(-1/2)=-1/2 \Rightarrow f(2012)=-1/2\) \)

CB
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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