Question Statement

Find the root of the equation $x{e}^x=2$ in the interval $[0,1]$ using the Regula Falsi method, correct to two decimal places.

---

Background and Explanation

The Regula Falsi method (or False Position method) is a numerical technique for finding roots that combines the bisection method's bracketing property with the secant method's faster convergence. It requires an interval $[a,b]$ where $f(a)$ and $f(b)$ have opposite signs, guaranteeing a root exists between them by the Intermediate Value Theorem.

---

Solution

Step 1: Define the function and verify the root exists

First, rewrite the equation as $f(x)=x{e}^x-2=0$.

Check the function values at the endpoints:

• At $x=0$: $f(0)=0\cdot e^0-2=-2<0$
• At $x=1$: $f(1)=1\cdot e^1-2=e-2\approx 2.7183-2=0.7183>0$

Since $f(0)<0$ and $f(1)>0$, and $f(x)$ is continuous on $[0,1]$, a root lies between $0$ and $1$.

Step 2: First Iteration ($x_0$)

Using the Regula Falsi formula with $a=0$ and $b=1$: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$

Substituting the values: $\begin{array}{rl}{x}_0& =\frac{0\cdot f(1)-1\cdot f(0)}{f(1)-f(0)}\\ & =\frac{0-(-2)}{0.7183-(-2)}\\ & =\frac{2}{2.7183}\\ & \approx 0.7358\end{array}$

Evaluate $f(x_0)$: $\begin{array}{rl}f(x_0)& =f(0.7358)=0.7358\cdot e^{0.7358}-2\\ & \approx -0.4643<0\end{array}$

Since $f(0.7358)<0$ and $f(1)>0$, the root lies in the updated interval $[0.7358,1]$.

Step 3: Second Iteration ($x_1$)

Apply the formula with $a=0.7358$ and $b=1$: $\begin{array}{rl}{x}_1& =\frac{0.7358\cdot f(1)-1\cdot f(0.7358)}{f(1)-f(0.7358)}\\ & =\frac{0.7358(0.7183)-(-0.4643)}{0.7183-(-0.4643)}\\ & =\frac{0.5285+0.4643}{1.1826}\\ & \approx 0.8395\end{array}$

Evaluate $f(x_1)$: $\begin{array}{rl}f(x_1)& =f(0.8395)=0.8395\cdot e^{0.8395}-2\\ & \approx -0.0564<0\end{array}$

Since $f(0.8395)<0$ and $f(1)>0$, the updated interval is $[0.8395,1]$.

Step 4: Third Iteration ($x_2$)

Apply the formula with $a=0.8395$ and $b=1$: $\begin{array}{rl}{x}_2& =\frac{0.8395\cdot f(1)-1\cdot f(0.8395)}{f(1)-f(0.8395)}\\ & =\frac{0.8395(0.7183)-(-0.0564)}{0.7183-(-0.0564)}\\ & =\frac{0.6030+0.0564}{0.7747}\\ & \approx 0.8512\end{array}$

Evaluate $f(x_2)$: $\begin{array}{rl}f(x_2)& =f(0.8512)=0.8512\cdot e^{0.8512}-2\\ & \approx -0.0061<0\end{array}$

Since $f(0.8512)<0$ and $f(1)>0$, the updated interval is $[0.8512,1]$.

Step 5: Fourth Iteration ($x_3$) and Convergence

Apply the formula with $a=0.8512$ and $b=1$: $\begin{array}{rl}{x}_3& =\frac{0.8512\cdot f(1)-1\cdot f(0.8512)}{f(1)-f(0.8512)}\\ & =\frac{0.8512(0.7183)-(-0.0061)}{0.7183-(-0.0061)}\\ & =\frac{0.6114+0.0061}{0.7244}\\ & \approx 0.8524\end{array}$

Conclusion

Comparing the last two approximations:

• $x_2\approx 0.8512$
• $x_3\approx 0.8524$

Up to two decimal places, both values round to $0.85$. Therefore, the root of the equation $x{e}^x=2$ correct to two decimal places is:

$\boxed{0.85}$

---

Key Formulas or Methods Used

• Function definition: $f(x)=x{e}^x-2$
• Regula Falsi formula: $x_n=\frac{a\cdot f(b)-b\cdot f(a)}{f(b)-f(a)}$
• Intermediate Value Theorem: If $f(a)\cdot f(b)<0$, then there exists at least one root in $(a,b)$
• Convergence criterion: Stop when successive approximations agree to the required number of decimal places (two decimal places in this case)

---

Summary of Steps

1. Define $f(x)=x{e}^x-2$ and verify $f(0)=-2<0$ and $f(1)\approx 0.7183>0$, confirming a root in $[0,1]$.
2. First iteration: Calculate $x_0=0.7358$ using Regula Falsi; since $f(x_0)<0$, update interval to $[0.7358,1]$.
3. Second iteration: Calculate $x_1=0.8395$; since $f(x_1)<0$, update interval to $[0.8395,1]$.
4. Third iteration: Calculate $x_2=0.8512$; since $f(x_2)<0$, update interval to $[0.8512,1]$.
5. Fourth iteration: Calculate $x_3=0.8524$.
6. Verify convergence: $x_2$ and $x_3$ agree to two decimal places ($0.85$), so the root is $0.85$.