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) Iff(x)=x1 anda=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 \(f(x)\) on the interval \([a, a + h]\) and then simplify the expression. Afterward, we will use this simplified expression to guess the instantaneous rate of change (IRC) or the derivative of the function at \(x = a\).

### Problem (a)

Given:
- \(f(x) = x^1 = x\)
- \(a = 2\)

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

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

Substitute \(f(x) = x\) and \(a = 2\):

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

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

\[
f(2+h) = 2+h, \quad f(2) = 2
\]

Substitute these values into the formula:

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

Simplify the expression:

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

Since \(h > 0\), we can safely cancel \(h\) in the numerator and the denominator:

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

### 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 \(f(x)\) evaluated at \(x = 2\). 

Since \(f(x) = x\) is a linear function, its derivative is constant:

\[
f'(x) = 1
\]

Thus, the instantaneous rate of change (the slope of the tangent line) at \(x = 2\) is:

\[
f_{\text{IRC}}(2) = f'(2) = 1
\]

### Conclusion

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

Would you like more details or have any questions?

#### 5 Related Questions:
1. How would you compute the average rate of change for a quadratic function?
2. What is the geometric interpretation of the average rate of change?
3. How does the average rate of change relate to the concept of the derivative?
4. Can the process used here be applied to non-linear functions, and how does it differ?
5. How do you find the slope of the tangent line to a function at any given point?

#### Tip:
To better understand the derivative, consider it as the limit of the average rate of change as the interval becomes infinitesimally small.

Learn how to compute the average rate of change and instantaneous rate of change for functions using a step-by-step approach. This problem involves understanding the concept of average rate of change over an interval and its connection to the derivative or instantaneous rate of change at a specific point. Suitable for high school students studying calculus.

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