# How to Use the Bisection Method: Practice Problems

$$\definecolor{importantColor}{RGB}{255,0,0} \definecolor{secondaryColor}{RGB}{255,0,255}$$

Step 1

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.

Step 2

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

$$\begin{array}{cccc|cc} &{\mbox{Finding the New Interval}}&&{\mbox{Next Approximation}} \\[8pt] f(\mbox{left}) & f(\mbox{mid}) & f(\mbox{right}) & \mbox{New Interval} & \mbox{Midpoint} & \mbox{Max Error}\\ \hline &&{\mbox{Starting Interval:}}& [1,2] & \blue{1.5} & \pm 0.5\\ f(1)=-6 & f(\red{1.5}) \approx -2 & f(\red 2) = 9 & [1.5, 2] & \blue{1.75} & \pm0.25\\ f(\red{1.5}) \approx -2 & f(\red{1.75})\approx 2.4 & f(2)=9 & [1.5,1.75] & \blue{1.625} & \pm0.125\\ f(1.5) \approx -2 & f(\red{1.625})\approx -0.03 & f(\red{1.75}) \approx 2.4 & [1.625,1.75] & \blue{1.6875} & \pm0.0625 \end{array}$$

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.

Step 1

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{array}{cl} x & {f(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{array}

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

Step 2

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

$$\begin{array}{cccc|cc} &&{\mbox{Finding the New Interval}}&&{\mbox{Next Approximation}} \\[8pt] f(\mbox{left}) & f(\mbox{mid}) & f(\mbox{right}) & \mbox{New Interval} & \mbox{Midpoint} & \mbox{Max Error}\\ \hline &&{\mbox{Starting Interval:}}& [3,4] & \blue{3.5} & \pm 0.5\\ f(\red 3)\approx -0.1 & f(\red{3.5}) \approx 0.1 & f(4) \approx 0.3 & [3, 3.5] & \blue{3.25} & \pm0.25\\ f(\red 3) \approx -0.1 & f(\red{3.25}) \approx 0.01 & f(3.5)\approx 0.1 & [3,3.25] & \blue{3.125} & \pm0.125 \end{array}$$

The function has a root at approximately $$x = \blue{3.125}$$ with a maximum possible error of $$\pm0.125$$ units.

Step 1

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.

Step 2

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.
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ %{}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -3 & f(\red{-3}) = 2\\ \mbox{Midpoint} & -2.5 & f(\red{-2.5}) \approx -0.8\\ \mbox{Current right-endpoint} & -2 & f(-2) = -3 \end{array}$$

• Second Interval: $$[-3,-2.5]$$
• Second Approximation: The midpoint of the second interval is $$x = -2.75$$.
• Associated Error: $$\pm0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ %{}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -3 & f(-3) = 2\\ \mbox{Midpoint} & -2.75 & f(\red{-2.75}) \approx 0.6\\ \mbox{Current right-endpoint} & -2.5 & f(\red{-2.5}) \approx -0.8 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ %{}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -2.75 & f(\red{-2.75}) \approx 0.6\\ \mbox{Midpoint} & -2.625 & f(\red{-2.625}) \approx -0.1\\ \mbox{Current right-endpoint} & -2.5 & f(-2.5) \approx -0.8 \end{array}$$

• 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.

Step 1

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
Step 2

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 2 & f(2) = 8\\ \mbox{Midpoint} & 2.5 & f(\red{2.5}) \approx 3\\ \mbox{Current right-endpoint} & 3 & f(\red 3) = -10 \end{array}$$

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

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 2.5 & f(\red{2.5}) \approx 3\\ \mbox{Midpoint} & 2.75 & f(\red{2.75}) \approx -2\\ \mbox{Current right-endpoint} & 3 & f(3) = -10 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 2.5 & f(2.5) \approx 3\\ \mbox{Midpoint} & 2.625 & f(\red{2.625}) \approx 0.9\\ \mbox{Current right-endpoint} & 2.75 & f(\red{2.75}) \approx -2 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 5th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 2.625 & f(\red{2.625}) \approx 0.9\\ \mbox{Midpoint} & 2.6875 & f(\red{2.6875}) \approx -0.5\\ \mbox{Current right-endpoint} & 2.75 & f(2.75) \approx -2 \end{array}$$

• 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.

Step 1

Identify the function by getting the equation equal to zero.

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

Step 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.
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ %{}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 0 & f(0) = -2\\ \mbox{Midpoint} & 1.5 & f(\red{1.5}) = -2.75\\ \mbox{Current right-endpoint} & 3 & f(\red 3) = 1 \end{array}$$

• 2nd Interval: $$[1.5,3]$$
• 2nd Approximation: $$x = 2.25$$
• Associated Error: $$\pm 0.75$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 1.5 & f(1.5) = -2.75\\ \mbox{Midpoint} & 2.25 & f(\red{2.25}) \approx -1.4\\ \mbox{Current right-endpoint} & 3 & f(\red 3) = 1 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 2.25 & f(2.25) = -1.4375\\ \mbox{Midpoint} & 2.625 & f(\red{2.625}) = -0.359375\\ \mbox{Current right-endpoint} & 3 & f(\red 3) = 1 \end{array}$$

• 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.

Step 1

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 $$f(x) = x^4 - x - 3$$.

Step 2

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

Step 3

Identify the 2nd interval, approximation and associated error.

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 0 & f(0) = -3\\ \mbox{Midpoint} & 1 & f(\red 1) = -3\\ \mbox{Current right-endpoint} & 2 & f(\red 2) = 11 \end{array}$$

• 2nd Interval: $$[1,2]$$
• 2nd Approximation: $$x = 1.5$$
• Associated Error $$\pm0.5$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 1 & f(\red 1) = -3\\ \mbox{Midpoint} & 1.5 & f(\red{1.5}) \approx 0.6\\ \mbox{Current right-endpoint} & 2 & f(2) = 11 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 1 & f(1) = -3\\ \mbox{Midpoint} & 1.25 & f(\red{1.25}) \approx -1.8\\ \mbox{Current right-endpoint} & 1.5 & f(\red{1.5}) \approx 0.6 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 5th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 1.25 & f(1.25) \approx -1.8\\ \mbox{Midpoint} & 1.375 & f(\red{1.375}) \approx -0.8\\ \mbox{Current right-endpoint} & 1.5 & f(\red{1.5}) \approx 0.6 \end{array}$$

• 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.

Step 1

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 $$f(x)= x^3 -9x^2 + 20x -13$$.

Step 2

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.

$$\begin{array}{c|c} x & f(x)\\ \hline 5 & -13\\ 6 & -1\\ 7 & 29 \end{array}$$

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
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 6 & f(\red 6) = -1\\ \mbox{Midpoint} & 6.5 & f(\red{6.5}) \approx 11.4\\ \mbox{Current right-endpoint} & 7 & f(7) = 29 \end{array}$$

• 2nd Interval: $$[6, 6.5]$$
• 2nd Approximation $$x = 6.25$$
• Associated Error: $$\pm 0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 6 & f(\red 6) = -1\\ \mbox{Midpoint} & 6.25 & f(\red{6.25}) \approx 4.6\\ \mbox{Current right-endpoint} & 6.5 & f(6.5) \approx 11.4 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 6 & f(\red 6) = -1\\ \mbox{Midpoint} & 6.125 & f(\red{6.125}) \approx 1.6\\ \mbox{Current right-endpoint} & 6.25 & f(6.25) \approx 4.6 \end{array}$$

• 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.

Step 1

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

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

Step 2

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.

$$\begin{array}{c|c} x & f(x)\\ \hline -1 & 2\\ -2 & 3\\ -3 & 2\\ -4 & -7\\ \end{array}$$

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

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -4 & f(-4) = -7\\ \mbox{Midpoint} & -3.5 & f(\red{-3.5}) \approx -1.1\\ \mbox{Current right-endpoint} & -3 & f(\red{-3}) = 2 \end{array}$$

• 2nd Interval: $$[-3.5, -3]$$
• 2nd Approximation: $$x = -3.25$$
• Associated Error: $$\pm0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -3.5 & f(\red{-3.5}) \approx -1.1\\ \mbox{Midpoint} & -3.25 & f(\red{-3.25}) \approx 0.7\\ \mbox{Current right-endpoint} & -3 & f(-3) = 2 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -3.5 & f(-3.5) \approx -1.1\\ \mbox{Midpoint} & -3.375 & f(\red{-3.375}) \approx -0.1\\ \mbox{Current right-endpoint} & -3.25 & f(\red{-3.25}) \approx 0.7 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 5th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & -3.375 & f(\red{-3.375}) \approx -0.1\\ \mbox{Midpoint} & -3.3125 & f(\red{-3.3125}) \approx 0.3\\ \mbox{Current right-endpoint} & -3.25 & f(-3.25) \approx 0.7 \end{array}$$

• 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.

Step 1

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

\\ \begin{align*} x & = \sqrt{71}\\ x^2 & = 71\\ x^2 -71 & =0 \end{align*} \\

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

Step 2

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

We know $$\sqrt{71}$$ 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
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 8 & f(\red 8) = -7\\ \mbox{Midpoint} & 8.5 & f(\red{8.5}) = 1.25\\ \mbox{Current right-endpoint} & 9 & f(9) = 10 \end{array}$$

• 2nd Interval: $$[8, 8.5]$$
• 2nd Approximation: $$x = 8.25$$
• Associated Error: $$\pm0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 8 & f(8) = -7\\ \mbox{Midpoint} & 8.25 & f(\red{8.25}) \approx -2.9\\ \mbox{Current right-endpoint} & 8.5 & f(\red{8.5}) = 1.25 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Left endpoint} & 8.25 & f(8.25) \approx -2.9\\ \mbox{Midpoint} & 8.375 & f(\red{8.375}) \approx -0.9\\ \mbox{Right endpoint} & 8.5 & f(\red{8.5}) = 1.25 \end{array}$$

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

$$\sqrt{71}\approx 8.4375$$ with a maximum error in this approximation of $$0.0625$$.

Step 1

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

\\ \begin{align*} x & = \frac 1 {\sqrt[5] 3}\\ x^5 & = \frac 1 3\\ 3x^5 & = 1\\ 3x^5 - 1 & = 0 \end{align*} \\

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

Step 2

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

Since $$0< \frac 1 {\sqrt[5] 3} < 1$$, we should be able to use $$[0,1]$$ as the first interval. A quick check of the function values confirms this.

$$\\ \begin{array}{c|c} x & f(x)\\ \hline 0 & -1\\ 1 & 2 \end{array} \\$$

• 1st Interval: $$[0,1]$$
• 1st Approximation: $$x = 0.5$$
• Associated Error: $$\pm0.5$$ units
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 0 & f(0) = -1\\ \mbox{Midpoint} & 0.5 & f(\red{0.5}) \approx -0.9\\ \mbox{Current right-endpoint} & 1 & f(\red 1) = 2 \end{array}$$

• 2nd Interval: $$[0.5, 1]$$
• 2nd Approximation: $$x=0.75$$
• Associated Error: $$\pm0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 0.5 & f(0.5) \approx -0.9\\ \mbox{Midpoint} & 0.75 & f(\red{0.75}) \approx -0.2\\ \mbox{Current right-endpoint} & 1 & f(\red 1) = 2 \end{array}$$

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.

Step 1

Find a nonlinear function with a root at $$\sqrt{125}$$.

\\ \begin{align*} x & = \sqrt{125}\\ x^2 & = 125\\ x^2 - 125 & = 0\\ \end{align*} \\

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

Step 2

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

Since $$11^2 = 121$$ and $$12^2 = 144$$ we know $$11 < \sqrt{125} < 12$$.

• First Interval: $$[11,12]$$
• First approximation: $$x = 11.5$$
• Associated Error: $$\pm0.5$$ units
Step 3

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

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 11 & f(\red{11}) =-4\\ \mbox{Midpoint} & 11.5 & f(\red{11.5}) = 7.25\\ \mbox{Current right-endpoint} & 12 & f(12) = 19 \end{array}$$

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

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

Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 11 & f(\red{11}) =-4\\ \mbox{Midpoint} & 11.25 & f(\red{11.25}) \approx 1.6\\ \mbox{Current right-endpoint} & 11.5 & f(11.5) = 7.25 \end{array}$$

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

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

$$\sqrt{125} \approx 11.125$$ with a maximum error of $$0.125$$ units.

Step 1

Find a nonlinear function with a root at $$\frac {\sqrt[4]{12500}} 2$$

\\ \begin{align*} x & = \frac{\sqrt[4]{12500}} 2\\ x^4 & = \frac{12500}{16}\\ x^4 & = \frac{3125} 4\\ 4x^4 & = 3125\\ 4x^4 - 3125 & = 0 \end{align*} \\

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

Step 2

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

Since $$10^4 = 10{,}000$$ is about the right size, we try $$f(10) = 36{,}875$$

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.

$$\\ \begin{array}{c|c} x & f(x)\\ \hline 5 & -625\\ 6 & 2059 \end{array} \\$$

Well, that was convenient.

• 1st Interval: $$[5,6]$$
• 1st Approximation: $$x = 5.5$$
• Associated Error: $$\pm0.5$$ units
Step 3

Find the second interval, approximation, and associated error.

$$\begin{array}{rc|l} {\mbox{Finding the 2nd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 5 & f(\red 5) =-625\\ \mbox{Midpoint} & 5.5 & f(\red{5.5}) = 535.25\\ \mbox{Current right-endpoint} & 6 & f(6) = 2059 \end{array}$$

• 2nd Interval: $$[5,5.5]$$
• 2nd Approximation: $$x = 5.25$$
• Associated Error: $$\pm0.25$$ units
Step 4

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

$$\begin{array}{rc|l} {\mbox{Finding the 3rd Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 5 & f(5) =-625\\ \mbox{Midpoint} & 5.25 & f(\red{5.25}) \approx -86.2\\ \mbox{Current right-endpoint} & 5.5 & f(\red{5.5}) = 535.25 \end{array}$$

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

$$\begin{array}{rc|l} {\mbox{Finding the 4th Interval}}\\ {} & x & f(x)\\ \hline \mbox{Current left-endpoint} & 5.25 & f(\red{5.25}) \approx -86.2\\ \mbox{Midpoint} & 5.375 & f(\red{5.375}) \approx 213.7\\ \mbox{Current right-endpoint} & 5.5 & f(5.5) = 535.25 \end{array}$$

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

$$\frac 1 2 \cdot \sqrt[4]{12500} \approx 5.3125$$ with a maximum error of 0.0625.

Step 1

Solve $$0.5^n(b-a)$$ for $$n$$ when $$a = 2$$ and $$b = 5$$

\\ \begin{align*} 0.5^n(5-2) & = 0.01\\ 0.5^n\cdot 3 & =\frac 1 {10}\\[6pt] \left(\frac 1 2\right)^n & = \frac 1 {30}\\[6pt] n\cdot \ln\left(\frac 1 2\right) & = \ln\left(\frac 1 {30}\right)\\[6pt] n\cdot\left(\ln 1 - \ln 2\right) & \ln 1 - \ln 30\\[6pt] -n\ln 2 & = - \ln 30\\[6pt] n & = \frac{\ln 30}{\ln 2}\\[6pt] & \approx 4.90732 \end{align*} \\

We will need to use at least 5 iterations in order to ensure the accuracy.

Step 1

Solve $$0.5^n(b-a) = 0.02$$ for $$n$$ when $$a = -1$$ and $$b = 1$$

\\ \begin{align*} 0.5^n\left(1-(-1)\right) & = 0.02\\[6pt] \left(\frac 1 2\right)^n\cdot 2 & = \frac 1 {50}\\[6pt] \left(\frac 1 2\right)^n & = \frac 1 {100}\\[6pt] n\ln\left(\frac 1 2\right) & = \ln\left(\frac 1 {100}\right)\\[6pt] n\left(\ln 1 - \ln 2\right) & = \ln 1 - \ln 100\\[6pt] -n\ln 2 & = -\ln 100\\[6pt] n & = \frac{\ln 100}{\ln 2}\\[6pt] & \approx 6.64473 \end{align*} \\