Question Statement

Find the root of the equation $3x-e^x=0$ in the interval $[0,1]$ correct to two decimal places using the Bisection method.

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Background and Explanation

The Bisection method is a root-finding algorithm that repeatedly bisects an interval and selects a subinterval in which a root must lie. It is based on the Intermediate Value Theorem, which guarantees that a continuous function with values of opposite signs at the endpoints of an interval must have at least one root within that interval.

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Solution

Define the function: $f(x)=3x-e^x$

Step 1: Verify the initial interval $[0,1]$ contains a root

Evaluate the function at the endpoints: $f(0)=3(0)-e^0=0-1=-1<0$

$f(1)=3(1)-e^1=3-e\approx 0.281>0$

Since $f(0)<0$ and $f(1)>0$, and $f(x)$ is continuous, there exists at least one root in the interval $[0,1]$.

Step 2: First Bisection

Calculate the midpoint: $x_0=\frac{0+1}{2}=0.5$

Evaluate the function: $f(x_0)=f(0.5)=3(0.5)-e^{0.5}=1.5-\mathrm{1.6487...}\approx -0.148<0$

Since $f(0.5)<0$ and $f(1)>0$, the root lies in the updated interval $[0.5,1]$.

Step 3: Second Bisection

$x_1=\frac{0.5+1}{2}=0.75$

$f(x_1)=f(0.75)=3(0.75)-e^{0.75}=2.25-\mathrm{2.117...}\approx 0.132>0$

Since $f(0.5)<0$ and $f(0.75)>0$, the updated interval is $[0.5,0.75]$.

Step 4: Third Bisection

$x_2=\frac{0.5+0.75}{2}=0.625$

$f(x_2)=f(0.625)=3(0.625)-e^{0.625}=1.875-\mathrm{1.868...}\approx 0.006>0$

The updated interval is $[0.5,0.625]$.

Step 5: Fourth Bisection

$x_3=\frac{0.5+0.625}{2}=0.5625$

$f(x_3)=f(0.5625)=3(0.5625)-e^{0.5625}=1.6875-\mathrm{1.754...}\approx -0.067<0$

The updated interval is $[0.5625,0.625]$.

Step 6: Fifth Bisection

$x_4=\frac{0.5625+0.625}{2}=0.59375$

$f(x_4)=f(0.59375)=3(0.59375)-e^{0.59375}=1.78125-\mathrm{1.810...}\approx -0.029<0$

The updated interval is $[0.59375,0.625]$.

Step 7: Sixth Bisection

$x_5=\frac{0.59375+0.625}{2}=0.609375$

$f(x_5)=f(0.609375)=3(0.609375)-e^{0.609375}=1.828125-\mathrm{1.839...}\approx -0.011<0$

The updated interval is $[0.609375,0.625]$.

Step 8: Seventh Bisection

$x_6=\frac{0.609375+0.625}{2}=0.6171875$

$f(x_6)=f(0.6171875)=3(0.6171875)-e^{0.6171875}=1.8515625-\mathrm{1.8537...}\approx -0.0021<0$

The updated interval is $[0.6171875,0.625]$.

Step 9: Eighth Bisection

$x_7=\frac{0.6171875+0.625}{2}=0.62109375$

$f(x_7)=f(0.62109375)=3(0.62109375)-e^{0.62109375}=1.86328125-\mathrm{1.8610...}\approx 0.0023>0$

The updated interval is $[0.6171875,0.62109375]$.

Step 10: Ninth Bisection

$x_8=\frac{0.6171875+0.62109375}{2}=0.619140625$

$f(x_8)=f(0.619140625)=3(0.619140625)-e^{0.619140625}=1.857421875-\mathrm{1.85733...}\approx 0.000090>0$

The updated interval is $[0.6171875,0.619140625]$.

Conclusion:

The interval endpoints are $0.6171875$ and $0.619140625$. When rounded to two decimal places, both values become $0.62$. Therefore, the root of the equation correct to two decimal places is:

$\boxed{0.62}$

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Key Formulas or Methods Used

• Intermediate Value Theorem: If $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$, then there exists $c\in (a,b)$ such that $f(c)=0$
• Bisection Formula: $x_n=\frac{a_n+b_n}{2}$
• Interval Update Rule:
  • If $f(x_n)<0$, new interval is $[x_n,b_n]$
  • If $f(x_n)>0$, new interval is $[a_n,x_n]$
• Stopping Criterion: When both endpoints of the interval round to the same value to the desired number of decimal places (two decimal places in this case)

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Summary of Steps

1. Define $f(x)=3x-e^x$ and verify $f(0)=-1<0$ and $f(1)\approx 0.281>0$ to confirm a root exists in $[0,1]$.
2. Iterate using the bisection formula $x_n=\frac{a+b}{2}$ and evaluate $f(x_n)$.
3. Update the interval: if $f(x_n)$ is negative, replace the left endpoint; if positive, replace the right endpoint.
4. Repeat the process for 9 iterations until the interval $[0.6171875,0.619140625]$ is obtained.
5. Verify that both endpoints round to $0.62$ to two decimal places, and conclude the root is $0.62$.