We keep the hypotheses and notation of question 3.2. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$. (a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that $$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$ (b) Compute $$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$ as a function of the derivatives of $f$.
We keep the hypotheses and notation of question 3.2. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$.\\
(a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that
$$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$
(b) Compute
$$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$
as a function of the derivatives of $f$.