Thread: Math Q&A Game
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AKG is offline
Dec13-05, 01:51 PM
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P: 2,589
siddharth, you asked:
Can you prove that the only possible solutions are f(x)=x and f(x)=0
I'm simply saying that f(x) = |x| satisfies the given criteria [i.e. |xy/2| = |x||y|/2 for all real x, y and f'(1) = 1 = |1| = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = |x| is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.