 Solution

Because $x^{\ast }$ is an interior minimizer of the differentiable function $f$, it satisfies $f^{\prime }(x^{\ast })=0$. Apply the Taylor theorem with Lagrange remainder to $f^{\prime }$ about $x_n$. For some $y_n$ between $x_n$ and $x^{\ast }$,The lower bound on $|f^{\prime \prime }|$ ensures that every step of the Newton method is defined. Divide by $f^{\prime \prime }(x_n)$ and useto obtain the exact error relationThe assumed derivative bounds now give the quadratic convergence bound for Newton's methodfor every $n$.