I tried to solve this problem (from a while ago):

https://brilliant.org/mathematics-problem/evaluating-a-function/?group=tUgjRZ0bXcg5

I, because of silly mistakes, used all three tries, and then wanted to view the solution. It said I had to spend points, so I did, but then the screen went blank and there was no solution. Even reloading now doesn't help.

Could you please help me either 1) view the solution on the actual page, or 2) help me get the solution?

Thanks, Michael

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## Comments

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TopNewestI can't figure out what I did wrong. Please help me.

(Spoilers!)

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So I got \(g(x+y) = g(x) + g(y) + xy\) as in the official solution (where \(g(t) = \dfrac{f(t)}{t+1}\).) Then I did \(x = y\) to get \(g(2x) = 2g(x) + x^2.\) After that I did \(y = 2x\) to get \[g(3x) = g(x) + g(2x) + 2x^2 = g(x) + [2g(x) + x^2] + 2x^2 \\ = 3g(x) + 3x^2.\] Sticking \(x = 1/3\) in we get \[g(1) = 3g(1/3) + 1/3\] or \[2 = 3g(1/3) + 1/3\] so \(g(1/3) = 5/9.\)

Then using the double formula we get

\[g(2/3) = 2g(1/3) + (1/3)^2 = 10/9 + 1/9 = 11/9.\]

Finally, plugging in \(y = 1\) and \(x = 2/3\) into the very first equation, we get

\[g(5/3) = g(2/3) + g(1) + 2/3 \\ = 11/9 + 2 + 2/3 = 35/9,\]

so \[f(5/3) = g(5/3) \cdot 3/8 = (35/9)(3/8) = 35/24.\]

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It is all okay until the final step. Note that \[g(t)(t+1)=f(t)\] not \[f(t)(t+1)=g(t)\] Then, \[f(5/3)=g(5/3)\cdot 8/3=(35/9)(8/3)=280/27\]

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Ohhhh.... of course I mess up the algebra at the very end. Thanks a lot!!

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Hi Michael,

I think we fixed it, at least it is no longer bugging out for me. For the last few days we've had some finicky solution displaying behavior, but it should be all sorted out now.

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Yes, it's fixed. Thanks so much!

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