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Let $a_1 \in \mathbb{R}$. Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$.

This question seems hard to solve since we don't know what $a_1$ is. We want both $|x|,|x-a_1|$ to be as small as possible or as large as possible to find the maximum or minimum, so how do I find where that is possible? Maybe it might occur at the median or some other point.

user19405892
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3 Answers3

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Set $b=a_1/2$ and the translation $x\mapsto x-b$, $y\mapsto y$. This leads to considering the function $$ f(x)=\frac{1}{1+|x-b|}+\frac{1}{1+|x+b|} $$ and it's not restrictive to assume $b\ge0$. Leave out the case $b=0$, for the moment. The function is even, so we just need to study it for $x\ge0$ and we can write it as $$ f(x)=\begin{cases} \dfrac{1}{1+b-x}+\dfrac{1}{1+x+b} & \text{if $0\le x<b$}\\[6px] \dfrac{1}{1+x-b}+\dfrac{1}{1+x+b} & \text{if $x\ge b$} \end{cases} $$ You can see that $f(0)=2/(1+b)$ and that $$ \lim_{x\to\infty}f(x)=0 $$ Thus there is no absolute minimum, because $f(x)>0$ for all $x$. We also have $$ f(b)=1+\frac{1}{1+2b} $$

The derivative is $$ f'(x)=\begin{cases} \dfrac{1}{(1+b-x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $0\le x<b$}\\[6px] -\dfrac{1}{(1-b+x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $x>b$} \end{cases} $$ Note that $f$ is decreasing in the interval $(b,\infty)$ (and not differentiable at $b$). In the interval $[0,b)$ we can write $$ f'(x)=\frac{4x(1+b)}{(1+b-x)^2(1+b+x)^2} $$ so the function has zero derivative at $0$ and is increasing in the interval $(0,b)$.

This should be enough to finish up.

The case $b=0$ is quite easy: $$ \frac{2}{1+|x|}\le 2 $$ for all $x$, equality only for $x=0$.

egreg
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Obviously $h(x) > 0$, with $h(x) \to 0$ as $x \to \pm \infty$. Thus the infimum is $0$, but there is no minimum.

Candidates for the maximum are the two points ($0$ and $a_1$) where the function is non-differentiable, plus any points where the derivative is $0$. However, the function is convex on intervals not containing $0$ and $a_1$, so points where the derivative is $0$ would be local minima, not maxima.

Robert Israel
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  • How do you know the function doesn't increase without bound? – user19405892 Mar 15 '16 at 17:53
  • "there is no minimum": are you sure ? –  Mar 15 '16 at 17:53
  • @user19405892: The function does not increase without bound because each term is $\le 1$. – André Nicolas Mar 15 '16 at 17:56
  • I think this is easier said then done. How can we find all points where the function is not differentiable and where it has a local maxima? – user19405892 Mar 15 '16 at 18:03
  • @user19405892: The denominators are never zero, and are differentiable everywhere except where the absolute values are zero, so $h$ is differentiable everywhere except $x=0$ and $x=a_1$. These are your likely candidates (indeed, in light of Robert Israel's remarks, your only candidates) for the locations of the maxima. Also notice that $h(0) = h(a_1)$. –  Mar 15 '16 at 18:18
  • @YvesDaoust Yes, I'm sure. There is a local minimum at $a_1/2$, but no global minimum. – Robert Israel Mar 15 '16 at 19:32
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WLOG, $a_1=2$.

The domain can be decomposed in three pieces:

$$x\le0\implies f(x)=\frac1{1-x}+\frac1{3-x},\\ 0\le x\le2\implies f(x)=\frac1{1+x}+\frac1{3-x}=\frac4{(1+x)(3-x)},\\ 2\le x\implies f(x)=\frac1{1+x}+\frac1{x-1}.$$

The first (magenta) and third (black) pieces have a derivative which is the sum of two squares and doesn't cancel. The second piece (light green) is the inverse of a downward parabola and has a local minimum at $(1,1)$.

The junction points form two maxima (angular points), at $(0,\frac43)$ and $(2,\frac43)$.

enter image description here

For other values of $a_1$, stretch the plot horizontally.