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

## 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 \[f(5/3) = g(5/3) \cdot 3/8 = (35/9)(3/8) = 35/24.\] – Michael Tang · 3 years, 8 months ago

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– Daniel Chiu · 3 years, 8 months ago

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

– Michael Tang · 3 years, 8 months ago

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

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. – Peter Taylor Staff · 3 years, 8 months ago

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– Michael Tang · 3 years, 8 months ago

Yes, it's fixed. Thanks so much!Log in to reply