Question Statement

A company models the profit $P(x)$ in thousands of dollars from producing $x$ units of a product with the following nonlinear profit function:

$P(x)=x^3-4x^2-7x+10$

The company wants to determine the production level $x$ where the profit is zero $(P(x)=0)$. Use the Bisection Method to approximate the root within the interval $[-1,1.5]$.

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Background and Explanation

The Bisection Method is a root-finding algorithm that works by repeatedly dividing an interval in half and selecting the subinterval where the function changes sign. It requires a continuous function and an initial interval $[a,b]$ where $f(a)$ and $f(b)$ have opposite signs, guaranteeing at least one root exists in the interval by the Intermediate Value Theorem.

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Solution

First, we verify that a root exists within $[-1,1.5]$ by evaluating the function at the endpoints:

$P(-1)=(-1{)}^3-4(-1{)}^2-7(-1)+10=-1-4+7+10=12>0$

$P(1.5)=(1.5{)}^3-4(1.5{)}^2-7(1.5)+10=3.375-9-10.5+10=-6.125<0$

Since $P(-1)>0$ and $P(1.5)<0$, and $P(x)$ is continuous (as a polynomial), there must be at least one root in the interval $[-1,1.5]$.

1st Iteration

Bisect the initial interval $[-1,1.5]$ to find the first midpoint:

$x_1=\frac{-1+1.5}{2}=0.25$

Evaluate the function at this point:

$f(x_1)=f(0.25)=(0.25{)}^3-4(0.25{)}^2-7(0.25)+10=8.0156>0$

Since $f(0.25)>0$ (same sign as $f(-1)$) and $f(1.5)<0$, the root must lie in the subinterval $[0.25,1.5]$.

2nd Iteration

Bisect the updated interval $[0.25,1.5]$:

$x_2=\frac{0.25+1.5}{2}=0.875$

Evaluate the function:

$f(x_2)=f(0.875)=(0.875{)}^3-4(0.875{)}^2-7(0.875)+10=1.4824>0$

Since $f(0.875)>0$ and $f(1.5)<0$, the root lies in the updated interval $[0.875,1.5]$.

3rd Iteration

Bisect the interval $[0.875,1.5]$:

$x_3=\frac{0.875+1.5}{2}=1.1875$

Evaluate the function:

$f(x_3)=f(1.1875)=(1.1875{)}^3-4(1.1875{)}^2-7(1.1875)+10=-2.278<0$

Since $f(0.875)>0$ and $f(1.1875)<0$, the root lies in the subinterval $[0.875,1.1875]$.

4th Iteration

Bisect the interval $[0.875,1.1875]$:

$x_4=\frac{0.875+1.1875}{2}=1.03125$

Therefore, after four iterations, the approximated root is:

$x\approx 1.03125$

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Key Formulas or Methods Used

• Bisection Method Formula: $x_n=\frac{a_{n-1}+b_{n-1}}{2}$ where $[a_{n-1},b_{n-1}]$ is the current interval containing the root
• Intermediate Value Theorem: If a continuous function $f$ has $f(a)$ and $f(b)$ with opposite signs on $[a,b]$, then there exists at least one root $c\in (a,b)$ where $f(c)=0$
• Interval Selection Rule: After calculating the midpoint $x_n$, compare the sign of $f(x_n)$ with the signs of the endpoints to determine which half contains the root

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Summary of Steps

1. Verify conditions: Check that $f(a)$ and $f(b)$ have opposite signs on the initial interval $[a,b]$
2. Calculate midpoint: Compute $x_n=\frac{a+b}{2}$
3. Evaluate function: Calculate $f(x_n)$ and determine its sign
4. Update interval:
   • If $f(x_n)$ has the same sign as $f(a)$, set the new interval to $[x_n,b]$
   • If $f(x_n)$ has the same sign as $f(b)$, set the new interval to $[a,x_n]$
5. Repeat: Continue bisecting until the desired precision is achieved or the maximum number of iterations is reached