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We are asked to calculate the difference quotient for the function \( f(x) = \sqrt{x} + 2 \) at \( x = 2 \), using three values for \( h \). The difference quotient formula is:

\[
\frac{f(2 + h) - f(2)}{h}
\]

### Step 1: Calculate \( f(2) \)

Since \( f(x) = \sqrt{x} + 2 \), we calculate \( f(2) \):

\[
f(2) = \sqrt{2} + 2
\]

We'll leave \( \sqrt{2} \) as an approximation later in the calculation.

### Step 2: Calculate \( f(2 + h) \) for different values of \( h \)

We need to compute this for three values: \( h = 0.1 \), \( h = 0.01 \), and \( h = -0.1 \). The general expression for \( f(2 + h) \) is:

\[
f(2 + h) = \sqrt{2 + h} + 2
\]

### Step 3: Compute the difference quotient for each \( h \)

Now, we calculate the difference quotient for each value of \( h \) by plugging \( f(2 + h) \) and \( f(2) \) into the difference quotient formula:

\[
\frac{f(2 + h) - f(2)}{h}
\]

### Step 4: Derivative estimate at \( x = 2 \)

Once we have the difference quotients, we estimate the derivative at \( x = 2 \). It is given that the derivative is of the form \( \frac{1}{n} \), so we will solve for \( n \).

I'll go ahead and calculate these values now.The calculations yield the following values for the difference quotient:

- For \( h = 0.1 \): \( \frac{f(2 + 0.1) - f(2)}{0.1} = 0.3492 \)
- For \( h = 0.01 \): \( \frac{f(2 + 0.01) - f(2)}{0.01} = 0.3531 \)
- For \( h = -0.1 \): \( \frac{f(2 - 0.1) - f(2)}{-0.1} = 0.3581 \)

### Step 5: Determine \( n \)
The derivative of \( f(x) \) at \( x = 2 \) seems to approach \( \frac{1}{n} \). From the difference quotient values, we see that the derivative is approximately 0.35. Therefore, \( n \) is likely close to 3 because \( \frac{1}{n} \approx 0.35 \).

Thus, \( n = 3 \).

Let me know if you need more details or further clarification!

Here are five related questions to explore:
1. How does the difference quotient relate to the derivative definition?
2. Why does the value of \( h \) affect the accuracy of the derivative approximation?
3. What is the geometrical interpretation of the derivative at \( x = 2 \)?
4. Can the derivative be calculated exactly using calculus techniques for \( f(x) = \sqrt{x} + 2 \)?
5. How does the function \( f(x) \) behave as \( x \) increases or decreases?

**Tip**: For small \( h \) values, the difference quotient gives a better approximation of the derivative!

Let f(x) = √x + 2. Calculate the difference quotient f(2 + h) − f(2) / h for h = 0.1, h = 0.01, and h = −0.1. If someone told you that the derivative (slope of the tangent line) of f(x) at x = 2 was 1/n for some integer n, what would you expect n to be?

Learn how to calculate the difference quotient for f(x) = √x + 2 at x = 2 with values of h = 0.1, 0.01, and -0.1. Understand how this relates to the derivative and the slope of the tangent line, and determine the value of n where the slope equals 1/n.

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