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Let's work through the problem step-by-step.

### Given function:
\[ f(x) = \frac{-2}{x^2 + 1} \]

### Part (a): Finding the First and Second Derivatives
1. **First derivative, \( f'(x) \):**
   Use the quotient rule for derivatives. If \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).

   Here, \( u = -2 \) and \( v = x^2 + 1 \).
   
   - \( u' = 0 \)
   - \( v' = 2x \)
   
   Applying the quotient rule:
   \[
   f'(x) = \frac{(0)(x^2 + 1) - (-2)(2x)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}
   \]

2. **Second derivative, \( f''(x) \):**
   Differentiate \( f'(x) = \frac{4x}{(x^2 + 1)^2} \) using the quotient rule again.

   - Let \( u = 4x \) and \( v = (x^2 + 1)^2 \).
   - \( u' = 4 \)
   - \( v' = 2(x^2 + 1)(2x) = 4x(x^2 + 1) \)

   Then:
   \[
   f''(x) = \frac{(4)(x^2 + 1)^2 - 4x \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}
   \]
   Simplifying:
   \[
   f''(x) = \frac{4(x^2 + 1 - x^2)}{(x^2 + 1)^3} = \frac{4}{(x^2 + 1)^3}
   \]

So, the answers for part (a) are:
- \( f'(x) = \frac{4x}{(x^2 + 1)^2} \)
- \( f''(x) = \frac{4(1 - x^2)}{(x^2 + 1)^3} \)

### Part (b): Identifying the Graph
The graph that displays \( f(x) \) in blue and \( f''(x) \) in red can be identified by observing the shapes of these functions. \( f(x) \) has a minimum point and behaves like an upside-down bell curve due to the \( \frac{-2}{x^2 + 1} \) form. The second derivative \( f''(x) \) will change signs, showing concavity intervals.

Based on typical shapes:
1. \( f(x) \) in blue should appear as a curve that dips.
2. \( f''(x) \) in red will have points of inflection where it crosses the x-axis.

Analyzing the graphs, **Graph A** appears to match this behavior, with the correct relative curvature.

### Part (c): Concavity Intervals
Using \( f''(x) = \frac{4(1 - x^2)}{(x^2 + 1)^3} \):

1. Set \( f''(x) = 0 \) to find inflection points:
   \[
   4(1 - x^2) = 0 \Rightarrow x = \pm 1
   \]

2. Test intervals around \( x = -1 \) and \( x = 1 \):
   - For \( x \in (-\infty, -1) \) and \( x \in (1, \infty) \), \( f''(x) < 0 \), so \( f \) is concave down.
   - For \( x \in (-1, 1) \), \( f''(x) > 0 \), so \( f \) is concave up.

Thus:
- \( f \) is concave up on \( (-1, 1) \).
- \( f \) is concave down on \( (-\infty, -1) \cup (1, \infty) \).

Would you like further details on any part of this solution, or do you have any specific questions?

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Here are 5 additional questions to explore:

1. How does the sign of \( f'(x) \) indicate the increasing and decreasing intervals of \( f(x) \)?
2. Why does the second derivative change sign at \( x = \pm 1 \)?
3. How would the graph of \( f(x) \) change if the function was positive instead of negative?
4. How can you confirm concavity by using the second derivative test?
5. Can you describe how to apply the first derivative test to find the extrema of \( f(x) \)?

**Tip:** Always check the intervals around critical points and inflection points to understand the behavior of the function fully.

Let f(x) = -2 / (x^2 + 1). a. Find the first and second derivatives. b. Identify the graph that displays f in blue and f'' in red. c. Using the graphs of f and f'', indicate where f is concave up and concave down.

Explore the solution for finding the first and second derivatives of f(x) = -2 / (x^2 + 1), identifying concavity intervals, and interpreting graph behavior with f(x) and f''(x) in calculus. Suitable for high school students in Grades 11-12.

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