Illustrate the construction of the secant method by means of a figure. When $f ^ { \prime } > 0$ on $I$, express $x _ { n + 1 }$ as a function of $x _ { n - 1 } , x _ { n }$ by means of the function $H _ { f }$ defined in question 3.3 of the third part. Recall: The secant method is defined as follows. Given $x_0, x_1 \in I$, for $n \geq 1$ we consider the line $L_n$ passing through the points $(x_{n-1}, f(x_{n-1}))$ and $(x_n, f(x_n))$ (with $L_n$ being the tangent line at $(x_n, f(x_n))$ when $x_n = x_{n-1}$). If $L_n$ intersects $\{(x,0) \mid x \in I\}$ at a unique point $(x,0)$, we define $x_{n+1} = x$.
Illustrate the construction of the secant method by means of a figure. When $f ^ { \prime } > 0$ on $I$, express $x _ { n + 1 }$ as a function of $x _ { n - 1 } , x _ { n }$ by means of the function $H _ { f }$ defined in question 3.3 of the third part.
Recall: The secant method is defined as follows. Given $x_0, x_1 \in I$, for $n \geq 1$ we consider the line $L_n$ passing through the points $(x_{n-1}, f(x_{n-1}))$ and $(x_n, f(x_n))$ (with $L_n$ being the tangent line at $(x_n, f(x_n))$ when $x_n = x_{n-1}$). If $L_n$ intersects $\{(x,0) \mid x \in I\}$ at a unique point $(x,0)$, we define $x_{n+1} = x$.