Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$

Question

Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$

$\bf{My\; Try::}$ For $\min$ of $f(x)$

$$\left(\sqrt{13-x}+\sqrt{x}\right)^2=13-x+x+2\sqrt{x}\sqrt{13-x}= 13+2\sqrt{x}\sqrt{13-x}\geq 13$$

Now $$\sqrt{x+27} + \sqrt{13-x}+\sqrt{x} \geq \sqrt{27} + \sqrt{13}$$

and equality hold at $x=0$

Now How can i calculate $\max$ of $f(x)\;,$ Help required, Thanks

Try to visualize these equations geometrically, use Pythogorean and triangle ineaulity. — Emre, May 31 '16 at 13:30
Do you want an "elegant" solution? Because this problem is not difficult at all with derivatives. — MathematicsStudent1122, May 31 '16 at 13:33
I didn't get how you found the minimum of the function. In particular, how did you conclude that $\sqrt{27}+\sqrt{13}$ is the infinum? It appears that instead of proving that no smaller value exists, you used a calculator to get the result or just overlooked, after reaching a certain minimum value. — Roby5, May 31 '16 at 13:37
@abiessu I know that but the OP's answers gives no reason for this. Still I prefer proofs which don't use this symmetric argument, as it could be a bit misleading at times. — Roby5, May 31 '16 at 13:52
To i also use same thing which abiessu mention., Using symmetry and condition of end point. — juantheron, May 31 '16 at 14:03
@juantheron Without that I suppose your conclusions appear a bit too loose. Kindly update your question. — Roby5, May 31 '16 at 14:04
@ Roby I also not mention domain of that function., above i have directly calculate min. of function, — juantheron, May 31 '16 at 14:06
@juantheron Domain is not very hard to see. But symmetry and extreme-value theorem are far-fetched. — Roby5, May 31 '16 at 14:07

score 3 · Accepted Answer · answered May 31 '16 at 13:43

3

Hint:by Cauchy-Schwarz inequality $$121=[(x+27)+3(13-x)+2x][1+\dfrac{1}{3}+\dfrac{1}{2}]\ge f^2(x)$$

answered May 31 '16 at 13:43

math110

1 Answers1