Question

Find the intervals of concavity and the points of inflexion of the function $f\left(x\right)={(x-1)}^{\frac{1}{3}}$

Answer 1

Given, $f\left(x\right)={(x-1)}^{\frac{1}{3}}$
Therefore, $f^{\prime }\left(x\right)=\frac{1}{3}{(x-1)}^{\frac{1}{3}-1}$
$=\frac{1}{3}{(x-1)}^{\frac{-2}{3}}$
and $f^{\prime \prime }\left(x\right)=\frac{-2}{9}{(x-1)}^{\frac{-5}{3}}$
when $x<1$, $f^{\prime \prime }\left(x\right)>0$
$f\left(x\right)$ is concave upward in the interval $(-\infty ,1)$
when $x>1$, $f^{\prime \prime }\left(x\right)<0$
$f\left(x\right)$ is concave downwards in the interval $(1,\infty )$
The curve changes from concave upward to concave downward when $x=1$.
The point of inflection is $[1,f\left(1\right)]$ (ie) $(1,0)$.