4. Assume that h > 0 and compute a simpliÖed expression for the average rate of change for each function f on the given interval [a; a + h]. Be sure to cancel the troublesome factor of h from the denominator in each case. This way, you can easily use the expression fARC[a;a+h] to guess the slope fIRC(a) of the tangent line at the point where x = a:
(a)  if f(x)=1/x and a=2,then fARC [2; 2 + h] =
and fIRC (2)

To solve this problem, we will compute the average rate of change (ARC) of the function \(f(x) = \frac{1}{x}\) over the interval \([a, a + h]\) and simplify the expression. Then, we will use this simplified expression to find the instantaneous rate of change (IRC), or the derivative, of the function at \(x = a = 2\).

### Step-by-Step Solution

Given:
- \(f(x) = \frac{1}{x}\)
- \(a = 2\)

The formula for the average rate of change of \(f(x)\) over the interval \([a, a + h]\) is:

\[
f_{\text{ARC}}[a, a + h] = \frac{f(a + h) - f(a)}{h}
\]

Substitute \(f(x) = \frac{1}{x}\) and \(a = 2\):

\[
f_{\text{ARC}}[2, 2 + h] = \frac{f(2 + h) - f(2)}{h}
\]

Compute \(f(2 + h)\) and \(f(2)\):

\[
f(2 + h) = \frac{1}{2 + h}, \quad f(2) = \frac{1}{2}
\]

Substitute these values into the formula:

\[
f_{\text{ARC}}[2, 2 + h] = \frac{\frac{1}{2 + h} - \frac{1}{2}}{h}
\]

### Simplify the Expression

First, find a common denominator for the fractions in the numerator:

\[
f_{\text{ARC}}[2, 2 + h] = \frac{\frac{2 - (2 + h)}{2(2 + h)}}{h}
\]

Simplify the numerator:

\[
f_{\text{ARC}}[2, 2 + h] = \frac{\frac{2 - 2 - h}{2(2 + h)}}{h} = \frac{\frac{-h}{2(2 + h)}}{h}
\]

Now, divide by \(h\):

\[
f_{\text{ARC}}[2, 2 + h] = \frac{-h}{h \cdot 2(2 + h)} = \frac{-1}{2(2 + h)}
\]

### Resulting Expression for the Average Rate of Change

\[
f_{\text{ARC}}[2, 2 + h] = \frac{-1}{2(2 + h)}
\]

### Guessing the Slope of the Tangent Line (Instantaneous Rate of Change)

The instantaneous rate of change of \(f(x)\) at \(x = 2\), denoted as \(f_{\text{IRC}}(2)\), is the derivative of \(f(x) = \frac{1}{x}\) evaluated at \(x = 2\). 

The derivative of \(f(x) = \frac{1}{x}\) is:

\[
f'(x) = -\frac{1}{x^2}
\]

Evaluate the derivative at \(x = 2\):

\[
f_{\text{IRC}}(2) = f'(2) = -\frac{1}{2^2} = -\frac{1}{4}
\]

### Conclusion

- The simplified expression for the average rate of change \(f_{\text{ARC}}[2, 2 + h]\) is \(\frac{-1}{2(2 + h)}\).
- The instantaneous rate of change \(f_{\text{IRC}}(2)\) (the slope of the tangent line at \(x = 2\)) is \(-\frac{1}{4}\).

Would you like more details or have any questions?

#### 5 Related Questions:
1. How does the process change when computing the derivative of more complex functions, such as polynomials?
2. How can the concept of limits be used to formally define the derivative?
3. What is the geometric interpretation of the derivative of a function at a point?
4. How do you find the average rate of change for a function over an arbitrary interval \([a, a + h]\)?
5. What are some practical applications of derivatives in real-world problems?

#### Tip:
Understanding the connection between average rate of change and instantaneous rate of change helps bridge the gap between basic algebra and more advanced calculus concepts, particularly the idea of limits.

Learn how to compute the average rate of change and the instantaneous rate of change (derivative) for the function f(x) = 1/x at x = 2. Detailed step-by-step solution provided with simplification of expressions and theorem applications. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation