A friend and I were messing around with an expression similar to the definition of $e$, but with an absolute value:


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?


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


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.