How to Use the Bisection Method:
Practice Problems

Find the 4th approximation of the positive root of the function $$f(x) = x^4 - 7$$ using the bisection method .

Since the function is continuous everywhere, find an appropriate starting interval.

Then, notice that $$f(1) = -6 < 0$$, but $$f(2) = 9 >0$$. Let's use $$[1, 2]$$ as the starting interval.

Set up and use the table of values as in the examples above. The approximations are in blue, the new intervals are in red.

The positive root of $$f(x) = x^4 - 7$$ is at approximately $$x = 1.6875$$. This approximation is off by at most $$\pm 0.0625$$ units.

Problem 2

Find the third approximation of the root of the function $$f(x) = \frac 1 2 x-\sqrt[3]$$ using the bisection method .

Since the function is continuous everywhere, determine an appropriate starting interval.

Set up a table of values to help us find an appropriate interval.

\begin x & \\ \hline 0 & f(0) = -1\\ 1 & f(1) \approx -0.8\\ 2 & f(2) \approx -0.4\\ 3 & f(3) \approx -0.1\\ 4 & f(4) \approx 0.3\\ \end

This table indicates the root is between $$x=3$$ and $$x = 4$$, so a good starting interval is $$[3,4]$$

Set up and use a table to track the appropriate values.

The function has a root at approximately $$x = \blue$$ with a maximum possible error of $$\pm0.125$$ units.

Problem 3

Approximate the negative root of the function $$f(x) = x^2-7$$ to within 0.1 of its actual value.

Since the function is continuous everywhere, determine an appropriate starting interval.

At $$x = -2$$ the function value is $$f(-2) = -3$$, and at $$x = -3$$ the function value is $$f(-3) = 2$$. We'll use $$[-3,-2]$$ as the starting interval.

Determine the first approximation and the maximum possible error in using it.

• 1st Approximation: The midpoint of the first interval is $$x = -2.5$$.
• Associated Error: $$\pm 0.5$$ units.

Determine the second interval, second approximation and the associated maximum error.

• Second Interval: $$[-3,-2.5]$$
• Second Approximation: The midpoint of the second interval is $$x = -2.75$$.
• Associated Error: $$\pm0.25$$ units

Repeat Step 3 with the new interval. Continue to repeat until the maximum error is less than $$0.1$$.

• Third interval: $$[-2.75,-2.5]$$.
• Third approximation: The midpoint is $$x = -2.625$$
• Associated error: $$\pm0.125$$ units.

• 4th interval: $$[-2.75,-2.625]$$.
• 4th approximation: Midpoint is at $$x = -2.6875$$
• Associated error: $$\pm 0.0625$$ units.

The negative root of the function is at approximately $$x = -2.6875$$ with a maximum error of only $$\pm0.0625$$ units.

Problem 4

Approximate the value of the root of $$f(x) = -3x^3+5x^2+14x-16$$ near $$x = 3$$ to within 0.05 of its actual value.

Determine an appropriate starting interval, the first approximation and its associated maximum error value.

First, notice that the function is continuous everywhere. Then, since we're told that the root is near $$x = 3$$ we can check that $$f(3) = -10$$.

Checking $$x = 4$$ we find that $$f(4) = -72$$, but at $$x = 2$$ the function value is $$f(2) = 8$$.

• First Interval: $$[2,3]$$
• First Approximation: The midpoint is at $$x = 2.5$$
• Associated error: $$\pm 0.5$$ units

Determine the second interval, the second approximation and the associated maximum error.

• Second Interval: $$[2.5,3]$$
• Second Approximation: The midpoint is at $$x = 2.75$$
• Associated Error: $$\pm0.25$$ units.

Repeat Step 2 until the maximum possible error is less than 0.05 units.

• Third interval: $$[2.5, 2.75]$$.
• Third approximation: The midpoint is at $$x = 2.625$$
• Associated error: $$\pm0.125$$ units.

• 4th interval: $$[2.625,2.75]$$
• 4th approximation: The midpoint is $$x = 2.6875$$.
• Associated error of $$\pm0.0625$$ units.

• 5th interval: $$[2.625, 2.6875]$$.
• 5th approximation: The midpoint is $$x = 2.65625$$
• Associated error: $$\pm0.03125$$ units.

The root of the function is approximately $$x = 2.65625$$ and has an associated maximum error of only $$\pm0.03125$$ units.

Problem 5

Find the 4th approximation to the solution of the equation below using the bisection method .

Identify the function by getting the equation equal to zero.

$$x^2 - 2x - 2 = 0$$
The function we'll use is $$f(x) = x^2 - 2x - 2$$.

Determine an appropriate starting interval, the first approximation and its associated maximum error.

At $$x = 0$$ the function value is $$f(0) = -2$$, while at $$x = 3$$ the function value is $$f(3) = 1$$.

• 1st Interval: $$[0,3]$$
• 1st Approximation: The midpoint is at $$x = 1.5$$.
• Associated Error: $$\pm1.5$$ units.

Determine the second interval, second approximation and its associated maximum error.

• 2nd Interval: $$[1.5,3]$$
• 2nd Approximation: $$x = 2.25$$
• Associated Error: $$\pm 0.75$$ units

Repeat Step 3 twice to complete the iterations of the bisection method for this question.

• 3rd Interval: $$[2.25,3]$$
• 3rd Approximation: $$x = 2.625$$
• Associated Error: $$\pm 0.375$$ units

• 4th Interval: $$[2.625,3]$$
• th Approximation: $$x=2.812$$
• Associated Error: $$\pm0.1875$$ units

The solution to the equation is approximately $$x =2.8125$$ with a maximum error of 0.1875 units.

Problem 6

Find the 5th approximation to the solution to the equation below, using the bisection method .

Identify the function we will use by rewriting the equation so it is set equal to zero.

$$x^4 - x -3 = 0$$
The function we will use is $$f(x) = x^4 - x - 3$$.

Identify the first interval, the first approximation and its associated maximum error.

Notice that at $$x = 0$$ the function value is $$f(0) = -3$$.

Also, at $$x = 2$$ the function value is $$f(2) = 11$$.

• 1st Interval: $$[0,2]$$
• 1st Approximation: $$x = 1$$
• Associated Error: $$\pm 1$$ unit

Identify the 2nd interval, approximation and associated error.

• 2nd Interval: $$[1,2]$$
• 2nd Approximation: $$x = 1.5$$
• Associated Error $$\pm0.5$$ units

Repeat Step 3 until you've found the 5th approximation.

• 3rd Interval: $$[1,1.5]$$
• 3rd Approximation: $$x = 1.25$$
• Associated Error: $$\pm0.25$$ units

• 4th Interval: $$[1.25, 1.5]$$
• 4th Approximation: $$x = 1.375$$
• Associated Error: $$\pm0.125$$ units

• 5th Interval: $$[1.375, 1.5]$$
• 5th Approximation: $$x = 1.4375$$
• Associated Error: $$\pm0.0625$$ units

The solution to the equation is approximately $$x = 1.4375$$. This approximation has an maximum error of at most 0.0625 units.

Problem 7

The equation below should have a solution that is larger than 5. Use the bisection method to approximate this solution to within 0.1 of its actual value.

$$x^3 + 18x -6 = 9x^2-2x+7$$

Identify the function we'll use by rewriting the equation so it is equal to zero.

$$x^3 -9x^2 + 20x -13 = 0$$
The function is $$f(x)= x^3 -9x^2 + 20x -13$$.

Determine the first interval, 1st approximation, and its associated error.

We know the solution is larger than 5, but we don't know how much larger. We set up a small table of values to help us out.

$$ \begin x & f(x)\\ \hline 5 & -13\\ 6 & -1\\ 7 & 29 \end $$

From this table we can select the first interval and determine the first approximation.

• 1st Interval: $$[6,7]$$
• 1st Approximation: $$x = 6.5$$
• Associated Error: $$\pm0.5$$ units

Find the second interval, second approximation and the associated error.

• 2nd Interval: $$[6, 6.5]$$
• 2nd Approximation $$x = 6.25$$
• Associated Error: $$\pm 0.25$$ units

Repeat Step 3 until the maximum error is less than the given tolerance of 0.1.

• 3rd Interval: $$[6,6.25]$$
• 3rd Approximation: $$x = 6.125$$
• Associated Error: $$\pm0.125$$ units

• 4th Interval: $$[6, 6.125]$$
• 4th Approximation $$x = 6.0625$$
• Associated Error: $$\pm0.0625$$ units

The solution to the equation is approximately $$x = 6.0625$$ with a maximum error of $$0.0625$$ units.

Problem 8

The only real solution to the equation below is negative. Approximate the value of this solution to within 0.05 units of its actual value.

$$x^3 + 7x^2 - 2 = 2x^2 - 7x -7$$

Rewrite the equation so we can identify the function we are working with.

$$x^3 + 5x^2 +7x +5 = 0$$
So, $$f(x) = x^3 + 5x^2 +7x +5$$

Identify the first interval, the first approximation, and the associated error.

We know the solution is negative, but that is all. Let's set up a table of values to get an idea of where our first interval should be.

$$ \begin x & f(x)\\ \hline -1 & 2\\ -2 & 3\\ -3 & 2\\ -4 & -7\\ \end $$

• 1st Interval: $$[-3,-4]$$
• 1st Approximation: $$x = -3.5$$
• Associated Error: $$\pm 0.5$$ units

Identify the 2nd interval, 2nd approximation and the associated maximum error.

• 2nd Interval: $$[-3.5, -3]$$
• 2nd Approximation: $$x = -3.25$$
• Associated Error: $$\pm0.25$$ units

Repeat Step 3 until the maximum error is less 0.05 units.

• 3rd Interval: $$[-3.5, -3.25]$$
• 3rd Approximation: $$x = -3.375$$
• Associated Error: $$\pm0.125$$

• 4th Interval: $$[-3.375, -3.25]$$
• 4th Approximation: $$x = -3.3125$$
• Associated Error: $$\pm0.0625$$ units

• 5th Interval: $$[-3.375,-3.3125]$$
• 5th Approximation: $$x = -3.34275$$
• Associated Error: $$\pm0.03125$$ units

The equation has a solution at approximately $$x = -3.34275$$ with a maximum error in the approximation of at most $$0.03125$$ units.

Problem 9

Use the bisection method to approximate the value of $$\sqrt$$. Find the 4th approximation.

Find a non-linear function whose root is at $$\sqrt 7$$

We'll use $$f(x) = x^2 - 71$$

Find the first interval, first approximation and its associated maximum error.

We know $$\sqrt$$ is larger than 8, but less than 9. We'll use $$[8,9]$$ as the first interval.

• 1st Interval: $$[8,9]$$
• 1st Approximation: $$x = 8.5$$
• Associated Error: $$\pm 0.5$$ units

Find the second interval, second approximation and the associated maximum error.

• 2nd Interval: $$[8, 8.5]$$
• 2nd Approximation: $$x = 8.25$$
• Associated Error: $$\pm0.25$$ units

Repeat Step 3 until you've found the 4th approximation.

• 3rd Interval: $$[8.25, 8.5]$$
• 3rd Approximation $$x = 8.375$$
• Associated Error: $$\pm0.125$$ units

• 4th Interval: $$[8.375, 8.5]$$
• 4th Approximation: $$x = 8.4375$$
• Associated Error: $$\pm0.0625$$ units

$$\sqrt\approx 8.4375$$ with a maximum error in this approximation of $$0.0625$$.

Problem 10

Use the bisection method to approximate the value of $$\frac 1 $$. Find the 3rd approximation.

Determine the nonlinear function we will use for the bisection method .

$$ \\ \begin x & = \frac 1 \\ x^5 & = \frac 1 3\\ 3x^5 & = 1\\ 3x^5 - 1 & = 0 \end \\ $$

We will use $$f(x) = 3x^5 - 1$$.

Find the first interval, first approximation and the associated error.

• 1st Interval: $$[0,1]$$
• 1st Approximation: $$x = 0.5$$
• Associated Error: $$\pm0.5$$ units

Determine the second interval, second approximation and the associated error.

• 2nd Interval: $$[0.5, 1]$$
• 2nd Approximation: $$x=0.75$$
• Associated Error: $$\pm0.25$$ units

Find the third interval, third approximation and its associated error.

Third Interval: $$[0.75,1]$$

Third Approximation: $$x = 0.875$$ with an error of 0.125 units.

$$\sqrt[5] 3\approx 0.875$$ with a maximum error of 0.125 units.

Problem 11

Use the bisection method to approximate the value of $$\sqrt$$ to within 0.125 units of the actual value.

Find a nonlinear function with a root at $$\sqrt$$.

$$ \\ \begin x & = \sqrt\\ x^2 & = 125\\ x^2 - 125 & = 0\\ \end \\ $$

We'll use $$f(x) = x^2 - 125$$.

Determine an appropriate starting interval. It's midpoint will be the first approximation.

Since $$11^2 = 121$$ and $$12^2 = 144$$ we know $$11 < \sqrt< 12$$.

• First Interval: $$[11,12]$$
• First approximation: $$x = 11.5$$
• Associated Error: $$\pm0.5$$ units

Determine the second interval, the second approximation, and the associated error value.

2nd Interval: $$[11,11.5]$$

2nd Approximation: $$x = 11.25$$ with a maximum error of 0.25 units.

Determine the third interval, the third approximation, and the associated error value.

Third Interval: $$[11,11.25]$$

Third Approximation: $$x = 11.125$$ with a maximum error of $$0.125$$.

$$\sqrt \approx 11.125$$ with a maximum error of $$0.125$$ units.

Problem 12

Use the bisection method to approximate the value of $$\frac > 2$$ to within 0.1 units of the actual value.

Find a nonlinear function with a root at $$\frac > 2$$

We'll use the function $$f(x) = 4x^4 - 3125$$

Determine the appropriate starting interval, the first approximation and the associated error.

Since $$10^4 = 10000$$ is about the right size, we try $$f(10) = 36875$$

By comparison, $$f(5) = -625$$, so the best starting interval is somewhere between $$x = 5$$ and $$x = 10$$. Let's make a table of values to help us narrow things down.

$$ \\ \begin x & f(x)\\ \hline 5 & -625\\ 6 & 2059 \end \\ $$

Well, that was convenient.

• 1st Interval: $$[5,6]$$
• 1st Approximation: $$x = 5.5$$
• Associated Error: $$\pm0.5$$ units

Find the second interval, approximation, and associated error.

• 2nd Interval: $$[5,5.5]$$
• 2nd Approximation: $$x = 5.25$$
• Associated Error: $$\pm0.25$$ units

Repeat Step 3 until the maximum error is smaller than the allowed tolerance.

• Third Interval: $$[5.25, 5.5]$$
• Third Approximation: The midpoint of the 3rd interval is $$x = 5.375$$
• Associated Error: $$\pm0.125$$ units

• Fourth Interval: $$[5.25, 5.375]$$
• Fourth Approximation: The midpoint of the 4th interval is $$x = 5.3125$$
• Associated Error: $$\pm0.0625$$ units

$$\frac 1 2 \cdot \sqrt[4] \approx 5.3125$$ with a maximum error of 0.0625.