What's the minimum of $(\frac{-1}{x}-1)^x$? A friend and I were messing around with an expression similar to the definition of $e$, but with an absolute value:
$$\big|\frac{1}{x}+1\big|^x$$
On the graphing calculator that there was a minimum around $(-0.218, 0.757)$.
We were trying to figure out the exact value of this minimum and came up with the expression,
$$\big(\frac{-1}{x}-1\big)^x$$
which is equivalent on the interval $(-1,0)$.
However, we quickly hit a wall here. Trying to solve using the derivative on a calculator created a humongous monstrosity that was too long to solve for zero.
What is the exact value of the minimum, and how could we calculate it?
 A: It becomes easier if you first make the substitution
$$t=-\frac1x-1$$
$$x=-\frac1{t+1}$$
to transform the function into the simpler
$$\large{f(t)=t^{-\frac1{t+1}}=e^{-\frac{\ln{(t)}}{t+1}}}$$
Which has a minimum where
$$g(t)=\frac{\ln(t)}{t+1}$$
is at a maximum. So we have
$$g'(t)=\frac{1+\frac1t-\ln{(t)}}{(t+1)^2}=0$$
$$1+\frac1t-\ln{(t)}=0$$
$$e^{1+\frac1t-\ln{(t)}}=1$$
$$\frac1te^{\frac1t}=\frac1e$$
$$\frac1t=W\left(\frac1e\right)$$
$$t=\frac1{W(\frac1e)}$$
Where $W(x)$ denotes the Lambert-W function. The associated $x$ value is
$$x=-\frac{W(\frac1e)}{W(\frac1e)+1}=\frac1{W(\frac1e)+1}-1$$
$$x\approx -0.217811705719800098779702924073255219818091596033700483129\dots$$
and the functions value is then
$$\large{\left(-\frac1x-1\right)^x=\left(W\left(\frac1e\right)\right)^{\frac{W(\frac1e)}{W(\frac1e)+1}}}$$
$$\left(-\frac1x-1\right)^x\approx0.756945106457583664584017088120241500061127660187365808210\dots$$
A: You can use implicit differentiation, much like the trick for differentiating $x^x$. You will get:
$$f'(x)=\left(-\left(1+\frac{1}{x}\right)\right)^x\left(-\frac{1}{x+1}+\ln\left(-\left(1+\frac{1}{x}\right)\right)\right).$$
This is zero when
$$\frac{1}{x+1}=\ln\left(-\left(1+\frac{1}{x}\right)\right).$$
As far as I know there is no closed-form solution to this equation (but you can check that $x\approx -0.218$ works).
