Calculus needed here?

Let $a=\sqrt [ 3 ]{ 1-x }$ and $b=\sqrt [ 3 ]{ x+3 }$ . Now $a+b=y$ . If each side is raised to the 3rd power, we have $a^3+b^3+3ab(a+b)=y^3$ . Substitute $a+b=y$ and $a^3+b^3 = 4$ to the left-hand-side, and finally obtain $ab=\frac{y^3-4}{3y}$ .

Since $a+b=y$ and $ab=\frac{y^3-4}{3y}$ , we note that $a$ and $b$ are roots to the equation $u^2-yu+\frac{y^3-4}{3y}=0$ . As $a$ and $b$ are real, the discriminant of the quadratic equation is non-negative, which means that $y^2 -4\left(\frac{y^3-4}{3y}\right)\ge 0$ . Now we have $0< y^3 \le 16$ . Thus $L=\sqrt [ 3 ]{ 16 } \approx 2.5198420$ . So $\lfloor 10000L\rfloor = 25198$ . Note that the value $L$ is obtained when $x=-1$ .

First we symmetrize the situation by defining $y = x+1$ . Then $f(y) = \sqrt[3]{2-y} + \sqrt[3]{2+y}.$ Calculus: $\frac{d}{dy}(2\pm y)^{1/3} = \pm\frac 1 3(2\pm y)^{-2/3}.$ If therefore $f'(y) = 0$ , we must have $(2+y)^{-2/3} = (2-y)^{-2/3}.$ This is the case if either $2+y = 2-y$ or if $2+y = -(2-y)$ , but the latter is impossible. We conclude that $y = 0$ , so that $L = 2\sqrt[3]{2} \approx 2.51984\dots$ Thus we post the answer as $\boxed{25198}$ .

To find the maximum value of the function, we first take the first derivative of the function (or the equation for the slope at each x value). $\frac{d}{dx} \sqrt [3]{1-x} + \sqrt [3]{x+3} = \frac{d}{dx} \sqrt[3]{1-x} + \frac{d}{dx} \sqrt[3]{x+3} = -\frac{(1-x)^\frac{-2}{3}}{3} +\frac{(x+3)^\frac{-2}{3}}{3}.$ In order to find the maximum point of this function, we would have to set the first derivative to $0.$ $-\frac{(1-x)^\frac{-2}{3}}{3} +\frac{(x+3)^\frac{-2}{3}}{3} = 0 \\ -\frac{(1-x)^\frac{-2}{3}}{3} = -\frac{(x+3)^\frac{-2}{3}}{3} \\ (x+3)^\frac{-2}{3} = (1-x)^\frac{-2}{3} \\ x+3 = 1-x \\ 2x = -2 \\ x = -1$ Now that we have an x position, we would have to check if it is the point of maximum value, minimum value, or an inflection point. Therefore, we find the second derivative of the function (the equation for the acceleration of the function at a given point). This turns out to be $-\frac{2(1-x)^\frac{-5}{3}}{9} - \frac{2(x+3)^\frac{-5}{3}}{9}$ We see that at $x=-2$ , the second derivative is positive and at $x=0$ , the second derivative is negative, therefore the maximum value of the function is at $x = -1$ Computing the y value gets us $\sqrt[3]{1-(-1))} + \sqrt[3]{(-1+3)} \\ \sqrt[3]{2} + \sqrt[3]{2} \\ 2 \cdot \sqrt[3]{2} \approx 2.5198421$ Now we compute $L = \lfloor2.5198421 \cdot 10000\rfloor = \boxed{25198}$

Relevant wiki: Power Mean Inequality (QAGH)

Disclaimer: This solution only works if we assumed the principal root of the cube root function. That is, we treat the domain of $\sqrt[3]{x}$ to be $x\geq 0$ , not $x\in \mathbb R$ .

Since $f(x)$ is real when $1-x \geq 0$ and $x+3\geq 0$ , then the domain of $f(x)$ is $-3 \leq x \leq 1$ .

Let $a = \sqrt[3]{1-x}$ and $b = \sqrt[3]{x+3}$ for $-3 \leq x \leq 1$ , then both $a$ and $b$ are non-negative numbers.

We can apply the power mean inequality ,

$\sqrt[3]{ \dfrac{a^3+b^3}2 } \geq \dfrac{a+b}2 \qquad \qquad \Leftrightarrow \qquad \qquad \dfrac42 \geq \left( \dfrac{f(x)}2 \right)^3 \qquad \qquad \Leftrightarrow \qquad \qquad f(x) \leq \sqrt[3]{16}.$

Equality occurs when $a = b \Leftrightarrow x = 2$ .

Hence, $L = \sqrt[3]{16} \Rightarrow \lfloor 10^4 \cdot L \rfloor = \boxed{25198}$ .

Let $g(x) = \sqrt[3]{1-x}$ and $h(x) = \sqrt[3]{x+3}$ . Note that $f(x) = g(x) + h(x)$ . And since $g(1-x) = h(x-3),$ their sum, $f(x)$ , is symmetric about the line $x=-1$ .

Now $g'(x) = \frac{-1}{3} (1-x)^{-2/3} < 0$ for all $x$ and $h'(x) = \frac{1}{3} (x+3)^{-2/3} > 0$ for all $x$ . We only care about the sign of $f' = g' + h'$ , so let's ignore the (1/3) for the moment.

Note that for $x < -1$ , $\left\|x+3\right\| < \left\|1-x\right\|$ , thus $(x+3)^{-2/3} > (1-x)^{-2/3},$ as exponentiation to a negative power is order reversing. Thus in this range, $h'(x) > -g'(x)$ , and $g'(x) + h'(x) > 0$ . Conversely, for $-1 < x , \left\|x+3\right\| > \left\|1-x\right\|$ , thus $(x+3)^{-2/3} < (1-x)^{-2/3}$ , which means $h'(x) < -g'(x)$ , and $g'(x) + h'(x) < 0$ .

Thus $f'(x) > 0$ for $x < -1$ and $f'(x) < 0$ for $x> -1$ , so a max occurs at $f(-1)$ . Evaluate $f(-1) = 2\sqrt[3]{2} = 2.5198\ldots$ , so the answer sought is $25198$ .

The possibility has been raised that $f(x)$ diverges as $x$ goes to infinity. This is not so. Let us consider the behavior of $f(x)$ as $x$ grows large. First, rewrite $f(x) = \sqrt[3]{x+3} - \sqrt[3]{x-1}.$ By the mean value theorem applied to the function $g(y) = y^{1/3},$ for large values of $x$ , there must be a value of $z(x)$ in the interval $[x-1,x+3]$ such that $f(x) = 4*g'(z(x)).$ Now $g'(y) = \frac{y^{-2/3}}{3} \rightarrow 0$ as $y \rightarrow \infty$ . Thus $f(x) \rightarrow 0$ as $x \rightarrow \infty$ . By symmetry, the same argument holds as $x \rightarrow -\infty$

This is my first time writing an actual algebraic solution for this type of problem so please critique me!

Using sum of two cubes identity, I rewrote $f(x)$ as

$f(x) = \dfrac{(\sqrt[3]{1-x} + \sqrt[3]{x+3})(\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} - \sqrt[3]{(1-x)(x+3)})}{\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} - \sqrt[3]{(1-x)(x+3)}}$

$= \dfrac{4}{\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} - \sqrt[3]{(1-x)(x+3)}} = \dfrac{4}{\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} + \sqrt[3]{(x-1)(x+3)}}$

I figured that I'd have to minimize the denominator (but only considering positive denominators) in order to maximize the numerator. We only consider positive denominators because if we were to consider a negative denominator, then consider when the denominator equal 1. Then $f(x)$ would equal $4$ , which is positive, and positive numbers are greater than negative numbers always, so dividing by a negative wouldn't give us a maximum value.

Looking at the denominator, I realized that $\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2}$ must always be positive since the innards of the cube roots are squared and thus positive, so $\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} \geqslant 0$ and $\sqrt[3]{(x-1)(x+3)}$ can be minimized if its innards are minimized and not between 0 and 1, or between -1 and 0. We know how to find the minimum of an upward opening parabola, which is its vertex, which occurs at $x = -1$ . So the minimum value of $\sqrt[3]{(1-x)^2} + \sqrt[3]{(x+3)^2} + \sqrt[3]{(x-1)(x+3)}$ should occur at $x = -1$ , and the maximum of $f(x)$ is $f(-1) = \boxed{ 2 \sqrt[3]{2} }$ .