Find the value of the derivative (if it exists) at the indicated extremum. (If an answer does not exist, enter DNE.)

*f*(*x*) = *x*2*/(x*2 + 2)

To determine the value of the derivative at the extremum of the function \( f(x) = \frac{x^2}{x^2 + 2} \), we can start by finding the derivative and then identifying any points where extrema might occur.

### Step 1: Differentiate \( f(x) \) using the Quotient Rule
Given:
\[
f(x) = \frac{x^2}{x^2 + 2}
\]
Let \( u = x^2 \) and \( v = x^2 + 2 \). Then:
\[
f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

Calculating \( u' \) and \( v' \):
- \( u' = 2x \)
- \( v' = 2x \)

Thus:
\[
f'(x) = \frac{(x^2 + 2)(2x) - (x^2)(2x)}{(x^2 + 2)^2}
\]
\[
= \frac{2x(x^2 + 2) - 2x(x^2)}{(x^2 + 2)^2}
\]
\[
= \frac{2x(x^2 + 2 - x^2)}{(x^2 + 2)^2}
\]
\[
= \frac{2x \cdot 2}{(x^2 + 2)^2}
\]
\[
= \frac{4x}{(x^2 + 2)^2}
\]

### Step 2: Find the Critical Points by Setting \( f'(x) = 0 \)
To find potential extrema, set \( f'(x) = 0 \):
\[
\frac{4x}{(x^2 + 2)^2} = 0
\]
The numerator \( 4x = 0 \) when \( x = 0 \).

### Step 3: Determine if \( x = 0 \) is an Extremum
To confirm whether \( x = 0 \) is an extremum, consider the second derivative test or examine the behavior of \( f(x) \) around \( x = 0 \):
- Substitute \( x = 0 \) into \( f(x) \):
  \[
  f(0) = \frac{0^2}{0^2 + 2} = 0
  \]

Given \( f(x) \) increases for \( x < 0 \) and decreases for \( x > 0 \), \( x = 0 \) is a **local minimum**.

### Step 4: Evaluate \( f'(x) \) at \( x = 0 \)
Since \( f'(0) = \frac{4 \cdot 0}{(0^2 + 2)^2} = 0 \), the value of the derivative at the extremum \( x = 0 \) is:
\[
\boxed{0}
\]

Would you like a detailed explanation of any step?

---

1. What is the significance of the second derivative in determining extrema?
2. How do you determine whether an extremum is a maximum or minimum?
3. How does the Quotient Rule apply to more complex functions?
4. How can critical points be verified to confirm extrema?
5. Why is the derivative zero at an extremum in certain cases?

**Tip:** For rational functions, ensure the denominator does not equal zero to confirm continuity at points of interest.

Find the value of the derivative (if it exists) at the indicated extremum. f(x) = x^2 / (x^2 + 2)

Learn to find the derivative value at the extremum of f(x) = x^2 / (x^2 + 2) using the Quotient Rule and analysis of critical points. This calculus problem covers derivative calculations and identifying extrema using the second derivative test. Suitable for Grades 11-12.

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