Let f(x) = x/(x2 + 1) . Find the limit of the average rate of
change of f(x) on the interval [2, c] as c approaches 2. Solve without using L'Hôpital's Rule

Certainly! Let's solve the problem without using L'Hôpital's Rule.

### Step 1: Write the formula for the average rate of change
The average rate of change of the function \( f(x) = \frac{x}{x^2 + 1} \) over the interval \([2, c]\) is given by:

\[
\text{Average Rate of Change} = \frac{f(c) - f(2)}{c - 2}
\]

### Step 2: Calculate \( f(2) \) and \( f(c) \)
First, let's compute \( f(2) \):

\[
f(2) = \frac{2}{2^2 + 1} = \frac{2}{4 + 1} = \frac{2}{5}
\]

Next, express \( f(c) \):

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

### Step 3: Set up the expression for the average rate of change
Substituting these into the average rate of change formula:

\[
\text{Average Rate of Change} = \frac{\frac{c}{c^2 + 1} - \frac{2}{5}}{c - 2}
\]

### Step 4: Simplify the expression
We need to simplify the expression in the numerator:

\[
\frac{c}{c^2 + 1} - \frac{2}{5} = \frac{5c - 2(c^2 + 1)}{5(c^2 + 1)} = \frac{5c - 2c^2 - 2}{5(c^2 + 1)}
\]

So, the expression for the average rate of change becomes:

\[
\frac{\frac{5c - 2c^2 - 2}{5(c^2 + 1)}}{c - 2} = \frac{5c - 2c^2 - 2}{5(c^2 + 1)(c - 2)}
\]

### Step 5: Factor the numerator
Notice that the numerator \( -2c^2 + 5c - 2 \) is a quadratic expression. We can factor it as follows:

\[
-2c^2 + 5c - 2 = -(2c^2 - 5c + 2)
\]

Let's factor \( 2c^2 - 5c + 2 \):

To factor this quadratic expression, we find two numbers that multiply to \( 2 \times 2 = 4 \) and add to \(-5\). These numbers are \(-4\) and \(-1\). So, we rewrite the quadratic as:

\[
2c^2 - 4c - c + 2 = (2c - 1)(c - 2)
\]

Thus, the numerator becomes:

\[
-2c^2 + 5c - 2 = -(2c - 1)(c - 2)
\]

### Step 6: Cancel the common factor
Now, substitute this back into the expression:

\[
\frac{-(2c - 1)(c - 2)}{5(c^2 + 1)(c - 2)}
\]

We can cancel the \((c - 2)\) term in the numerator and denominator:

\[
\frac{-(2c - 1)}{5(c^2 + 1)}
\]

### Step 7: Take the limit as \( c \) approaches 2
Finally, we compute the limit as \( c \) approaches 2:

\[
\lim_{c \to 2} \frac{-(2c - 1)}{5(c^2 + 1)} = \frac{-(2(2) - 1)}{5(2^2 + 1)} = \frac{-(4 - 1)}{5(4 + 1)} = \frac{-3}{25}
\]

Thus, the limit is:

\[
\boxed{-\frac{3}{25}}
\]

This is the limit of the average rate of change of \( f(x) \) on the interval \([2, c]\) as \( c \) approaches 2.

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How do you factor different types of quadratic expressions?
2. What are the conditions where cancellation of terms is valid?
3. How does factoring help in simplifying limits?
4. What are other techniques to find limits without using L'Hôpital's Rule?
5. How does the limit of the average rate of change relate to the derivative?

**Tip:** When simplifying expressions, always check for common factors that can be canceled to avoid indeterminate forms.

Discover how to find the limit of the average rate of change of f(x) = x/(x^2 + 1) on the interval [2, c] as c approaches 2. Learn the step-by-step solution without using L'Hôpital's Rule, involving limits, average rate of change, and quadratic functions. Suitable for advanced high school students.

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