Location of the maximum of the function $x \mapsto \frac{x^3}{e^x-1}$

Question

Determine location of the maximum of the function $$x \mapsto \frac{x^3}{e^x-1}$$ for $x>0$. Evaluate it with $2$-digit precision. Hint: $3 e^{-3} \approx 0.15$

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I calculated

$$\frac{{\rm d}}{{\rm d} x} \left( \frac{x^3}{e^x-1} \right) = \frac{3x^2(e^x-1)-x^3e^x}{(e^x-1)^2}$$

So the root can be calculated from equation $e^x(3-x)=3$. However, I don't know how to derive $x$ from this equation and I don't know how to use information from the hint

$$\frac{3}{e^3} \approx 0.15$$

Answer 1

The Lambert W function defines $W(z)$ to be the value $w$ that satisfies $we^w=z$.

Your equation can be written as $(x-3)e^{x-3} = -3/e^3$ so $x=W(-3/e^3) + 3$. Do you have some way of numerically approximating the Lambert W function?

Answer 2

Rearranging you have $x = 3(1 - e^{-x})$. In this case a simple fixed point iteration will do the job: $x' \leftarrow 3(1 - e^{-x})$ with an initial guess of $x = 1$ gets you 2.8214... in less than a dozen iterations.[Added] But see the comment of @g.kov.