Problem of the week #99 - February 17th, 2014

[anemone]

Evaluate $g(\sqrt[4]{2014})$ for a given function $g(x)=\sqrt[3]{\dfrac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}}+\sqrt[3]{\dfrac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}}$.

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

 

[anemone]

Congratulations to the following members for their correct solutions:

1. mente oscura
2. lfdahl
3. Opalg
4. MarkFL
5. magneto

Solution by Opalg:

Spoiler

The only plausible way to handle that horrendous-looking function $g$ would be if the expressions under the cube root signs are actually equal to the cube of something simpler, which looks as though it would have to be of the form $ax + b\sqrt{x^2 - 4}$ for some constants $a$, $b$. So the most promising thing is to look at $$\begin{aligned}\bigl( ax + b\sqrt{x^2 - 4}\bigr)^3 &= a^3x^3 + 3a^2bx^2\sqrt{x^2 - 4} + 3ab^2x(x^2-4) + b^3(x^2-4)\sqrt{x^2 - 4} \\ &= (a^3 + 3ab^2)x^3 - 12ab^2x + \bigl((3a^2b+b^3)x^2 - 4b^3\bigr)\sqrt{x^2 - 4}.\end{aligned}$$ If we compare the coefficients in that expression with those in $\dfrac{x^3-3x + (x^2-4)\sqrt{x^2 - 4}}2$, we get the conditions $$a^3+3ab^2 = \tfrac12, \qquad -12ab^2 = -\tfrac32,\qquad 3a^2b+b^3 = \tfrac12, \qquad -4b^3 = -\tfrac12.$$ The last of those conditions gives $b^3 = \frac18$, or $b = \frac12$, and the other three conditions are then all satisfied if we take $a = \frac12$ also. Therefore $$\sqrt[3]{\frac{x^3-3x + (x^2-4)\sqrt{x^2 - 4}}2} = \frac{x + \sqrt{x^2-4}}2.$$ If instead of the positive square root $\sqrt{x^2-4}$ we take the negative root $-\sqrt{x^2-4}$, then we get $$\sqrt[3]{\frac{x^3-3x - (x^2-4)\sqrt{x^2 - 4}}2} = \frac{x - \sqrt{x^2-4}}2.$$ Now add the two cube roots together to get $g(x) = \dfrac{x + \sqrt{x^2-4}}2 + \dfrac{x - \sqrt{x^2-4}}2 = x.$ So $g(x) = x$, and in particular $g\bigl(\sqrt[4]{2014}\bigr) = \sqrt[4]{2014} \approx 6.699.$

Solution by MarkFL:

Spoiler

Let:

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

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

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:

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