1. **Problem Statement:** Explain the methods to find roots of equations and solve two example problems using these methods. 2. **Mathematical Background:** Root-finding methods aim to find $x$ such that $f(x)=0$ for a given function $f$. 3. **Bisection Method:** Uses interval halving. If $f(a)$ and $f(b)$ have opposite signs, root lies in $(a,b)$. Formula: midpoint $c=\frac{a+b}{2}$. Replace $a$ or $b$ with $c$ based on sign. 4. **Regula Falsi (False Position) Method:** Uses a secant line between $(a,f(a))$ and $(b,f(b))$ to approximate root. Formula: $$c = b - \frac{f(b)(b - a)}{f(b) - f(a)}$$ 5. **Newton-Raphson (NR) Method:** Uses tangent line at $x_n$ to find next approximation. Formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ Requires derivative $f'(x)$. 6. **Secant Method:** Similar to NR but uses two previous points to approximate derivative. Formula: $$x_{n+1} = x_n - f(x_n) \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$$ 7. **Successive Approximation (Fixed Point Iteration):** Rewrite $f(x)=0$ as $x=g(x)$ and iterate $x_{n+1} = g(x_n)$ until convergence. --- **Example 1: Find root of $f(x) = x^2 - 4$ using Bisection Method in $[1,3]$** 1. Check signs: $f(1) = -3$, $f(3) = 5$, opposite signs, root in $[1,3]$. 2. Midpoint $c=\frac{1+3}{2}=2$, $f(2)=0$, root found exactly. **Example 2: Find root of $f(x) = x^3 - x - 2$ using Newton-Raphson starting at $x_0=1.5$** 1. Compute $f(1.5) = 1.5^3 - 1.5 - 2 = 3.375 - 1.5 - 2 = -0.125$. 2. Compute derivative $f'(x) = 3x^2 - 1$, so $f'(1.5) = 3(2.25) - 1 = 6.75 - 1 = 5.75$. 3. Next approximation: $$x_1 = 1.5 - \frac{-0.125}{5.75} = 1.5 + 0.0217 = 1.5217$$ 4. Repeat until desired accuracy. Final answers: - Example 1 root: $x=2$ - Example 2 root approx: $x \approx 1.5217$ after one iteration.