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Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$

How to solve this problem without using derivatives?


Michael Rozenberg's edit:

Because with a derivative it's not so easy: $$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$ $$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$ $$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$ which gives $x_{max}=3.$

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    $\begingroup$ Tthe limit is minus infinity when $x\to\infty$. Thus, there is no minimum. $\endgroup$ – Wlod AA Dec 26 '19 at 4:48
  • $\begingroup$ @ Wlod AA So what is the maximum value of $A$? $\endgroup$ – Success Dec 26 '19 at 4:51
  • $\begingroup$ Ooops, I've misread the Q. :) $\endgroup$ – Wlod AA Dec 26 '19 at 4:54
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    $\begingroup$ Derivatives are really the best way to go in maximization and minimization problems $\endgroup$ – Zarrax Dec 26 '19 at 5:08
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    $\begingroup$ @MichaelRozenberg Sometimes, with some difficulty/cleverness you can find elementary solutions to such problems, but differentiation is a systematic approach that will work on a large range of problems, and won't require any tricks specific to a given problem. $\endgroup$ – Zarrax Dec 27 '19 at 19:29
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By C-S and AM-GM we obtain: $$\sqrt{x-2}+2\sqrt{x+1}-x+2019=\sqrt{x-2}+4\sqrt{\frac{x+1}{4}}-x+2019\leq$$ $$\leq\sqrt{(1+4)\left(x-2+4\cdot\frac{x+1}{4}\right)}-x+2019=\sqrt{5(2x-1)}-x+2019\leq$$ $$\leq\frac{5+2x-1}{2}-x+2019=2021.$$ The equality occurs for $x=3,$ which says that we got a maximal value.

The minimum does not exist. Try $x\rightarrow+\infty.$

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    $\begingroup$ A very good solution. Thank you very much! $\endgroup$ – Success Dec 26 '19 at 7:16
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    $\begingroup$ @Success You are welcome! $\endgroup$ – Michael Rozenberg Dec 26 '19 at 7:32
  • $\begingroup$ Can down-voter explain us why did you do it? $\endgroup$ – Michael Rozenberg Jan 2 at 18:43
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Let $f(x) = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$ then

$$ f'(x) = \frac{1}{2\sqrt{x - 2}} + \frac{1}{\sqrt{x+1}} - 1 $$

Equating this to zero for real $x$, we get $x = 3$. The second derivative is negative hence the global maximum is $f(3) = 2021.$

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    $\begingroup$ If not using the derivative of the function, how is this problem solved? $\endgroup$ – Success Dec 26 '19 at 5:03
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    $\begingroup$ @Success Well in that case, you should first mention in your question that you want to maximum without using derivative $\endgroup$ – Nilotpal Kanti Sinha Dec 26 '19 at 5:05
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    $\begingroup$ @Nilotpal Kanti Sinha, one tag is "algebra-pre-calculus". So, the banning of derivatives seems clear enough. $\endgroup$ – Bernard Massé Dec 26 '19 at 5:07
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    $\begingroup$ It's easier to see and say that the first derivative is decreasing for $\,x>2.\ $ To look at the 2nd derivative would be a pain in the neck. $\endgroup$ – Wlod AA Dec 26 '19 at 5:09
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    $\begingroup$ @BernardMassé That's OK but if you want something specific, mentioning it in the question makes it less likely to be over-sighted $\endgroup$ – Nilotpal Kanti Sinha Dec 26 '19 at 5:10
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$$ \forall_{y\ge -1}\quad\sqrt{1+y}\ \le\ 1+\frac y2 $$

hence

$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ =\ \sqrt{1+(x-3)}\ \le\ 1 +\frac{x-3}2 $$ and $$ \forall_{x\ge -1}\quad\sqrt{x+1}\ =\ 2\cdot\sqrt{1+\frac{x-3}4}\ \le \ 2+\frac{x-3}4 $$ hence $$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ \le\ 2+x $$ i.e. $$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ -\ x\ \le\ 2 $$

where the equality holds only for $\ x=3.\ $ Thus A (see the OP Question) attains its maximum $A=2021$ at $x=3$.   Great!

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    $\begingroup$ Nice; you could say $\forall y\color{red}\ge-1$ and $\forall x\color{red}\ge2$, right? $\endgroup$ – J. W. Tanner Dec 26 '19 at 19:12
  • $\begingroup$ @J.W.Tanner, right! Thank you -- I'll implement your improvement. $\endgroup$ – Wlod AA Dec 27 '19 at 0:01

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