inequality problem

**dimeola** · April 11th 2009, 10:27 PM

Hi all,

I can't find the answer for this question...

Find the region of x such that arcsin (x ) < x + (x^3)/6 holds.

I try differentiate x + (x^3)/6 - arcsin (x) and want to find the roots but it didn't work...

Please help.

**NonCommAlg** · April 12th 2009, 12:14 AM

Originally Posted by dimeola

Hi all,

I can't find the answer for this question...

Find the region of x such that arcsin (x ) < x + (x^3)/6 holds.

I try differentiate x + (x^3)/6 - arcsin (x) and want to find the roots but it didn't work...

Please help.

put $x=\sin t$ and $f(t)=t - \sin t - \frac{\sin^3 t}{6}, \ \ \frac{-\pi}{2} \leq t \leq \frac{\pi}{2}.$ see that $f(0)=0, \ f(-\pi/2) < 0, \ f(\pi/2) > 0.$ we're looking for values of $t$ for which $f(t) < 0.$

we have: $f'(t)=1 - \cos t - \frac{\cos t \sin ^2 t}{2}=\frac{(1- \cos t)^2(2 + \cos t)}{2} \geq 0.$ you should be able to see now that $-1 \leq x < 0$ is the answer.

**dimeola** · April 12th 2009, 07:39 AM

Hi,

Thank you very much

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