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- Feb 14, 2012

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- Thread starter
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- #1

- Feb 14, 2012

- 3,917

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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1. mente oscura

2. lfdahl

3. Opalg

4. MarkFL

5. magneto

Solution by Opalg:

Solution by MarkFL:

\(\displaystyle g_1(x)=\sqrt[3]{\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}}\)

\(\displaystyle g_2(x)=\sqrt[3]{\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}}\)

Thus, we have:

\(\displaystyle g(x)=g_1(x)+g_2(x)\)

Cubing both sides, we obtain:

\(\displaystyle g^3(x)=g_1^3(x)+3g_1^2(x)g_2(x)+3g_1(x)g_2^2(x)+g_2^3(x)\)

We may arrange this as:

\(\displaystyle g^3(x)=g_1^3(x)+g_2^3(x)+3g_1(x)g_2(x)\left(g_1(x)+g_2(x) \right)\)

Now, we find:

\(\displaystyle g_1^3(x)+g_2^3(x)=\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}=x^3-3x\)

\(\displaystyle g_1(x)g_2(x)=\sqrt[3]{\frac{\left(x^3-3x \right)^2-\left(x^2-1 \right)^2\left(x^2-4 \right)}{4}}=\)

\(\displaystyle \sqrt[3]{\frac{x^6-6x^4+9x^2-\left(x^4-6x^4+9x^2-4 \right)}{4}}=1\)

And given:

\(\displaystyle g(x)=g_1(x)+g_2(x)\)

We may then state:

\(\displaystyle g^3(x)=x^3-3x+3g(x)\)

\(\displaystyle g^3(x)-3g(x)=x^3-3x\)

Observing that the domain of $g$ is $2\le x$, we may then conclude that on this domain, we must have:

\(\displaystyle g(x)=x\)

Hence:

\(\displaystyle g\left(\sqrt[4]{2014} \right)=\sqrt[4]{2014}\)

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