For the function f left parenthesis x right parenthesis equals x squared​, compute the following average rates of change.
​(a) From 4 to 5
​(b) From 4 to 4.5
​(c) From 4 to 4.1
​(d) From 4 to 4.01
​(e) From 4 to 4.001
​(f) Use a graphing utility to graph each of the secant lines along with f.  What do you think is happening to the secant​ lines?
​(g) To what number are the slopes of the secant lines getting​ closer?

We are asked to compute the **average rates of change** of the function \( f(x) = x^2 \) between different intervals and analyze the behavior of secant lines.

### Average Rate of Change Formula:
The average rate of change of a function \( f(x) \) from \( x = a \) to \( x = b \) is given by:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

Where \( f(x) = x^2 \).

---

### (a) From \( 4 \) to \( 5 \):
We have \( a = 4 \) and \( b = 5 \).

\[
f(4) = 4^2 = 16
\]
\[
f(5) = 5^2 = 25
\]
\[
\text{Average Rate of Change} = \frac{f(5) - f(4)}{5 - 4} = \frac{25 - 16}{1} = 9
\]

---

### (b) From \( 4 \) to \( 4.5 \):
We have \( a = 4 \) and \( b = 4.5 \).

\[
f(4) = 16
\]
\[
f(4.5) = (4.5)^2 = 20.25
\]
\[
\text{Average Rate of Change} = \frac{f(4.5) - f(4)}{4.5 - 4} = \frac{20.25 - 16}{0.5} = \frac{4.25}{0.5} = 8.5
\]

---

### (c) From \( 4 \) to \( 4.1 \):
We have \( a = 4 \) and \( b = 4.1 \).

\[
f(4) = 16
\]
\[
f(4.1) = (4.1)^2 = 16.81
\]
\[
\text{Average Rate of Change} = \frac{f(4.1) - f(4)}{4.1 - 4} = \frac{16.81 - 16}{0.1} = \frac{0.81}{0.1} = 8.1
\]

---

### (d) From \( 4 \) to \( 4.01 \):
We have \( a = 4 \) and \( b = 4.01 \).

\[
f(4) = 16
\]
\[
f(4.01) = (4.01)^2 = 16.0801
\]
\[
\text{Average Rate of Change} = \frac{f(4.01) - f(4)}{4.01 - 4} = \frac{16.0801 - 16}{0.01} = \frac{0.0801}{0.01} = 8.01
\]

---

### (e) From \( 4 \) to \( 4.001 \):
We have \( a = 4 \) and \( b = 4.001 \).

\[
f(4) = 16
\]
\[
f(4.001) = (4.001)^2 = 16.008001
\]
\[
\text{Average Rate of Change} = \frac{f(4.001) - f(4)}{4.001 - 4} = \frac{16.008001 - 16}{0.001} = \frac{0.008001}{0.001} = 8.001
\]

---

### (f) Graphing the Secant Lines:
If you graph the function \( f(x) = x^2 \) along with the secant lines calculated above, you will notice that the secant lines become closer and closer to the tangent line at \( x = 4 \) as the second point ( \( b \)) approaches 4. 

### (g) To what number are the slopes of the secant lines getting closer?
As the points get closer to \( x = 4 \), the average rate of change approaches the **slope of the tangent line** at \( x = 4 \), which is the derivative of \( f(x) = x^2 \) at that point:

\[
f'(x) = 2x
\]
At \( x = 4 \):
\[
f'(4) = 2(4) = 8
\]
So, the slopes of the secant lines are getting closer to 8.

---

Would you like further explanation on any part of this problem?

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**Related Questions:**
1. How do you compute the average rate of change for a general function?
2. What is the geometric interpretation of a secant line?
3. How do secant lines relate to the concept of a derivative?
4. What happens to secant lines as the two points get infinitely close?
5. How do we formally define the derivative using the limit process?

**Tip:** The secant line's slope gives the average rate of change, while the tangent line's slope represents the instantaneous rate of change.

For the function f(x) = x^2, compute the following average rates of change. (a) From 4 to 5 (b) From 4 to 4.5 (c) From 4 to 4.1 (d) From 4 to 4.01 (e) From 4 to 4.001 (f) Use a graphing utility to graph each of the secant lines along with f. What do you think is happening to the secant lines? (g) To what number are the slopes of the secant lines getting closer?

Compute the average rates of change for the function f(x) = x^2 over various intervals and explore how secant lines approach the tangent line at x = 4. Analyze the behavior of the secant lines and understand how their slopes approach the derivative, f'(x) = 2x.

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