5

Find the maximum and minimum value of
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$.

given that $-1\le x\le 1$

I have solved the problem but i am just curious to know if there are any other ways to solve this particular problem other than the method i used below.

By using the fact that $\arcsin \left(x\right)+\arccos \left(x\right)$ =$\frac{\pi }{2}$

i found that

$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$=$3\left(\frac{\pi }{2}\right)^2\left(\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2+\frac{\left(8\pi -3\pi ^2\right)}{48}\right)$

so it is minimum when $\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2$$=0$

or $x=\sin \left(\frac{\pi }{4}\right)$

Therefore minimum value$=\frac{1}{32}\pi ^3$

and it is maximum when $\arcsin \left(x\right)=-\frac{\pi }{2}$

Therefore maximum value=$\frac{7}{8}\pi ^3$

Adesh
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2 Answers2

2

Over the interval $[-1,1]$ the functions $\arcsin(x)$ and $\arccos(x)$ have a constant sum, $\frac{\pi}{2}$.
It follows that the problem is equivalent to finding the minimum and maximum value of $$ g(t) = t^3+\left(\frac{\pi}{2}-t\right)^3 $$ over the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. By setting $t=\frac{\pi}{2}u$ this boils down to finding the minimum and maximum value of $h(u)=u^3+(1-u)^3 = 1-3u+3u^2$ over the interval $[-1,1]$. That is straightforward: the minimum lies at $u=\frac{1}{2}$ and the maximum lies at $u=-1$. By performing the inverse substitutions, the maximum of the original function is $\frac{7\pi^3}{8}$, attained at $x=-1$, and the minimum is $\frac{\pi^3}{32}$, attained at $x=\frac{1}{\sqrt{2}}$.

Jack D'Aurizio
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0

Help: a^3+b^3=(a+b)((a+b)^2-3ab)

Let p=Pi

ASinx+ACosx=p/2

f(x)=(ASinx)^3+(ACosx)^3 =p/2 (p^2/4-3.ASinx.ACosx) =p/2 (p^2/4-3.(p/2).ASinx+3.(ASinx)^2)

This shall be maximum for ASinx= -(p/2) Hence max value of the function will be:

f (x)max=(7/8) (p^2)

skgarg70g
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    Please use Mathjax to format your math text. This link may be a useful reference: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Dave Jun 26 '17 at 16:17
  • Is this not what they have already done, just with worse formatting and an apparent typo? – siegehalver Jun 26 '17 at 16:36