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 $\displaystyle x=\sin t$ and $\displaystyle f(t)=t - \sin t - \frac{\sin^3 t}{6}, \ \ \frac{-\pi}{2} \leq t \leq \frac{\pi}{2}.$ see that $\displaystyle f(0)=0, \ f(-\pi/2) < 0, \ f(\pi/2) > 0.$ we're looking for values of $\displaystyle t$ for which $\displaystyle f(t) < 0.$
we have: $\displaystyle f'(t)=1 - \cos t - \frac{\cos t \sin ^2 t}{2}=\frac{(1- \cos t)^2(2 + \cos t)}{2} \geq 0.$ you should be able to see now that $\displaystyle -1 \leq x < 0$ is the answer.