\begin{tabular}{|c|c|}
\hline
\begin{tabular}{l}
Step 3
\end{tabular} & [tex]$\frac{5+6}{2}=\frac{11}{2}$[/tex] \\
\hline
\begin{tabular}{l}
Step 4
\end{tabular} & \begin{tabular}{l}
Evaluate the rewritten equation at [tex]$x=\frac{11}{2}$[/tex]. \\
[tex]$\sqrt[3]{\left(\frac{11}{2}\right)+1}-\sqrt{\left(\frac{11}{2}\right)-2} \approx-0.0046$[/tex]
\end{tabular} \\
\hline
Step 5 & \begin{tabular}{l}
Since this value of [tex]$x$[/tex] yields a negative difference, use it to replace \\
the previous upper bound that yielded a negative difference, \\
[tex]$x=6$[/tex]. \\
So, the bounds are now [tex]$x=5$[/tex] and [tex]$x=\frac{11}{2}$[/tex].
\end{tabular} \\
\hline
\end{tabular}

In what step, if any, did George make his first mistake?

A. In step 1, George should have subtracted in the opposite order to set the equation equal to 0.

B. George made no mistakes, and his work is correct.

C. In step 5, George should have used [tex]$x=\frac{11}{2}$[/tex] as the new lower bound.

D. In step 3, George should have averaged the two differences instead of the two bounds.



Answer :

Let's walk through George's steps to determine if he made any mistakes:

### Step 3:
George calculates the midpoint of the interval [tex]\([5, 6]\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{5 + 6}{2} = \frac{11}{2} = 5.5 \][/tex]

### Step 4:
George evaluates the function at [tex]\( x = 5.5 \)[/tex]:
[tex]\[ \sqrt[3]{(5.5 + 1)} - \sqrt{(5.5 - 2)} \approx -0.0046 \][/tex]

### Step 5:
Based on the result of the function at midpoint, which is approximately [tex]\(-0.0046\)[/tex] (a negative value), George decides to replace the upper bound. This is because by his rule:
- If the result is positive, he would replace the lower bound.
- If the result is negative, he would replace the upper bound.

Since the result is negative, George updates the interval to:
[tex]\[ [5, 5.5] \][/tex]

### Answer Choices:
A. In step 1, George should have subtracted in the opposite order to set the equation equal to 0.
This doesn't seem relevant to the steps he took, so this choice is not applicable here.

B. George made no mistakes, and his work is correct.
Given the steps and rationale detailed, George has correctly followed the method without making any mistakes up to this point.

C. In step 5, George should have used [tex]\( x = \frac{11}{2} \)[/tex] as the new lower bound.
This would contradict his rule that a negative result dictates replacement of the upper bound.

D. In step 3, George should have averaged the two differences instead of the two bounds.
Averaging the differences isn't a part of the typical procedure and wouldn't apply to finding the midpoint of the interval.

Based on the clear and correct logic George applied at each step, the conclusion is that:

George made no mistakes, and his work is correct.

Thus, the correct answer is:

B. George made no mistakes, and his work is correct.