From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$. For all $x \in \left[ x _ { 0 } , 1 \right]$, we define $$h ( x ) = \sqrt { \left| f ( x ) - f \left( x _ { 0 } \right) \right| }$$ (a) Show that the function $h$ defines a bijection from $\left[ x _ { 0 } , 1 \right]$ to $[ 0 , h ( 1 ) ]$. (b) Show that the map $h$ is differentiable at $x _ { 0 }$ on the right, and that $h ^ { \prime } \left( x _ { 0 } \right) = \sqrt { \frac { f ^ { \prime \prime } \left( x _ { 0 } \right) } { 2 } }$.
From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$.
For all $x \in \left[ x _ { 0 } , 1 \right]$, we define
$$h ( x ) = \sqrt { \left| f ( x ) - f \left( x _ { 0 } \right) \right| }$$
(a) Show that the function $h$ defines a bijection from $\left[ x _ { 0 } , 1 \right]$ to $[ 0 , h ( 1 ) ]$.
(b) Show that the map $h$ is differentiable at $x _ { 0 }$ on the right, and that $h ^ { \prime } \left( x _ { 0 } \right) = \sqrt { \frac { f ^ { \prime \prime } \left( x _ { 0 } \right) } { 2 } }$.