2016/109) Let $-1 <=p <=1$. Show that the equation $4x^3-3x-p=0$ has a unique root in the interval $[\frac{1}{2}, 1]$ find it.

because p is between -1 and 1 so $cos^{-1}p$ is defined. let $x = \cos\,t$ so we get
$\cos\, 3t = p$
and hence $3t = \arccos p$ and so $x = \frac{1}{3}\cos^{-}p$
if $x > 1$ we get $4x^3-3x$ > 1 and if $x < \frac{1}{2}$ we get $4x^3 - 3x < -1$ and out side the
limit so x is between $\frac{1}{2}$ and 1