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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?

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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 $$\large{\left(-\frac1x-1\right)^x=\left(W\left(\frac1e\right)\right)^{\frac{W(\frac1e)}{W(\frac1e)+1}}}$$ $$\left(-\frac1x-1\right)^x\approx0.756945106457583664584017088120241500061127660187365808210\dots$$

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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).

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  • $\begingroup$ Should read $1/(x+1)$ instead of $-1(x+1)$ in the last equation. $\endgroup$ – Michael Hoppe Apr 12 at 15:20
  • $\begingroup$ Thank you, fixed. $\endgroup$ – kccu Apr 12 at 15:21
  • $\begingroup$ There is a closed form solution for this. See my answer. $\endgroup$ – Peter Foreman Apr 12 at 15:30
  • $\begingroup$ Yes, this is in fact the global minimum of $\left|\frac1x+1\right|^x$ $\endgroup$ – Peter Foreman Apr 12 at 15:51
  • $\begingroup$ The OP wants to find the minimum of $f(x) = \lvert \frac{1}{x}+1 \rvert ^x$ which is not defined at $x = 0$ and $x = -1$. We have $f(x) = f_1(x) = (-\frac{1}{x}-1)^x$ on $(-1,0)$ and $f(x) = f_2(x) = (\frac{1}{x}+1)^x$ else. So $f_1(1) = -2$, but it is irrelevant here. $\endgroup$ – Paul Frost Apr 12 at 16:17

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