5. Find the difference quotient for 𝑓(𝑥) =
1
𝑥
2

𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
 12
Find 𝑚𝑎𝑣𝑔 𝑓𝑜𝑟 [−1, 2]

Let's break down the problem into two parts.

### Part 1: Difference Quotient for \( f(x) = \frac{1}{x^2} \)

The difference quotient formula is given as:

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

For \( f(x) = \frac{1}{x^2} \), let's first calculate \( f(x+h) \).

\[
f(x+h) = \frac{1}{(x+h)^2}
\]

Now, apply the difference quotient formula:

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

We need to combine the two terms in the numerator over a common denominator. The common denominator is \( (x+h)^2 x^2 \), so:

\[
\frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{(x+h)^2 x^2}
\]

Now, expand \( (x+h)^2 \):

\[
(x+h)^2 = x^2 + 2xh + h^2
\]

So the numerator becomes:

\[
x^2 - (x^2 + 2xh + h^2) = -2xh - h^2
\]

Substitute this back into the difference quotient formula:

\[
\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2}{h(x+h)^2 x^2}
\]

Simplify by canceling \( h \) in the numerator and denominator:

\[
\frac{-2xh - h^2}{h(x+h)^2 x^2} = \frac{-2x - h}{(x+h)^2 x^2}
\]

So the difference quotient is:

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

### Part 2: Find the Average Rate of Change (M_avg) of \( f(x) \) on the interval \([-1, 2]\)

The average rate of change formula is:

\[
M_{\text{avg}} = \frac{f(b) - f(a)}{b - a}
\]

where \( a = -1 \) and \( b = 2 \), and \( f(x) = \frac{1}{x^2} \).

1. Calculate \( f(-1) \):

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

2. Calculate \( f(2) \):

\[
f(2) = \frac{1}{2^2} = \frac{1}{4}
\]

Now, use the average rate of change formula:

\[
M_{\text{avg}} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{\frac{1}{4} - 1}{2 + 1} = \frac{\frac{1}{4} - \frac{4}{4}}{3} = \frac{-\frac{3}{4}}{3} = -\frac{1}{4}
\]

Thus, the average rate of change of \( f(x) \) on the interval \([-1, 2]\) is:

\[
M_{\text{avg}} = -\frac{1}{4}
\]

### Summary:
- **Difference quotient** for \( f(x) = \frac{1}{x^2} \): \( \frac{-2x - h}{(x+h)^2 x^2} \)
- **Average rate of change** on \([-1, 2]\): \( -\frac{1}{4} \)

Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the difference quotient relate to the derivative?
2. What is the significance of the average rate of change?
3. How would the difference quotient change for a linear function?
4. Can you generalize the process for finding the difference quotient for any rational function?
5. How does the average rate of change differ from instantaneous rate of change?

**Tip:** The difference quotient is essentially the definition of the derivative before you take the limit as \( h \to 0 \).

Find the difference quotient for f(x) = 1/x^2 and find the average rate of change of f(x) on the interval [-1, 2].

Learn how to find the difference quotient for the rational function f(x) = 1/x^2 and calculate the average rate of change on the interval [-1, 2]. This problem covers advanced algebraic concepts, suitable for students in Grades 10-12.

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