# Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$

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 $$x_{max}=3.$$

• Tthe limit is minus infinity when $x\to\infty$. Thus, there is no minimum. – Wlod AA Dec 26 '19 at 4:48
• @ Wlod AA So what is the maximum value of $A$? – Success Dec 26 '19 at 4:51
• Ooops, I've misread the Q. :) – Wlod AA Dec 26 '19 at 4:54
• Derivatives are really the best way to go in maximization and minimization problems – Zarrax Dec 26 '19 at 5:08
• @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. – Zarrax Dec 27 '19 at 19:29

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 $$x\rightarrow+\infty.$$

• A very good solution. Thank you very much! – Success Dec 26 '19 at 7:16
• @Success You are welcome! – Michael Rozenberg Dec 26 '19 at 7:32
• Can down-voter explain us why did you do it? – Michael Rozenberg Jan 2 at 18:43

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 $$f(3) = 2021.$$

• If not using the derivative of the function, how is this problem solved? – Success Dec 26 '19 at 5:03
• @Success Well in that case, you should first mention in your question that you want to maximum without using derivative – Nilotpal Kanti Sinha Dec 26 '19 at 5:05
• @Nilotpal Kanti Sinha, one tag is "algebra-pre-calculus". So, the banning of derivatives seems clear enough. – Bernard Massé Dec 26 '19 at 5:07
• 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. – Wlod AA Dec 26 '19 at 5:09
• @BernardMassé That's OK but if you want something specific, mentioning it in the question makes it less likely to be over-sighted – Nilotpal Kanti Sinha Dec 26 '19 at 5:10

$$\forall_{y\ge -1}\quad\sqrt{1+y}\ \le\ 1+\frac y2$$

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!

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