From the image, I can see a problem regarding a polynomial function $f(x) = x^4 - 32x^2 + 3$, with parts (a), (b), and (c) addressing increasing/decreasing intervals, local extrema, and inflection points/concavity. Here's an overview of what the problem entails and how it's solved:

Part (a): Increasing and Decreasing Intervals

The function's derivative, $f'(x) = 4x^3 - 64x$, determines where $f(x)$ is increasing or decreasing:

1. Critical Points: Solve $f'(x) = 0$, yielding $x = -4, 0, 4$.
2. Analyze $f'(x)$ sign changes in the intervals: $(-\infty, -4)$, $(-4, 0)$, $(0, 4)$, and $(4, \infty)$.
   • $f'(x) > 0$ (increasing) in $(-4, 0)$ and $(4, \infty)$.
   • $f'(x) < 0$ (decreasing) in $(-\infty, -4)$ and $(0, 4)$.

Part (b): Local Extrema

Evaluate $f(x)$ at the critical points $x = -4, 0, 4$:

• $f(-4) = -253$, a local minimum.
• $f(0) = 3$, a local maximum.
• $f(4) = -253$, a local minimum.

Part (c): Inflection Points and Concavity

The second derivative, $f''(x) = 12x^2 - 64$, helps analyze concavity:

1. Inflection Points: Solve $f''(x) = 0$, yielding $x = \pm \frac{4}{\sqrt{3}}$.
   • Evaluate $f(x)$ at $x = \pm \frac{4}{\sqrt{3}}$: the $y$-coordinates for the inflection points need re-calculating.
2. Concavity Intervals:
   • $f''(x) > 0$ (concave up) in $(-\infty, -\frac{4}{\sqrt{3}}) \cup (\frac{4}{\sqrt{3}}, \infty)$.
   • $f''(x) < 0$ (concave down) in $(-\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}})$.

Additional Questions:

1. How is the derivative used to classify increasing or decreasing intervals?
2. Why does $x = -4$ give a local minimum instead of a maximum?
3. What role do the inflection points play in determining the graph's shape?
4. Can you generalize the method to find extrema and concavity for any polynomial?
5. How do you interpret the intervals of concave up/down geometrically?

Tip: Always cross-check your derivative solutions by substituting back into the original function to verify critical points and inflection behavior.