How to Use the Bisection Method:
Practice Problems

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

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

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.

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 $$[3,4]$$

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

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

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.

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

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

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

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

• 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.

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

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

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

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

• 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.

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

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.

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

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

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

• 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.

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

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

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

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

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

• 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.

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

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

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

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

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

• 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.

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

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

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

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

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

• 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.

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

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

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

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

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

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

• 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 $$0.0625$$.

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

• 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.

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

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

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

We will need at least 7 iterations before the error tolerance is reached.