To solve this problem, let's go through each part systematically.

Given: $f(x) = \sin(\pi x) - \cos(\pi x)$ on the interval $[-1, 1]$.

We need to answer four parts:

1. Intervals of increase/decrease.
2. Local minima and maxima.
3. Intervals of concavity.
4. Inflection points.

Step 1: First Derivative for Increasing/Decreasing Intervals

To find where the function is increasing or decreasing, we need $f'(x)$.

$f'(x) = \pi \cos(\pi x) + \pi \sin(\pi x)$ Simplifying: $f'(x) = \pi (\cos(\pi x) + \sin(\pi x))$

Set $f'(x) = 0$ to find critical points:

$\cos(\pi x) + \sin(\pi x) = 0$ or $\sin(\pi x) = -\cos(\pi x)$

Dividing both sides by $\cos(\pi x)$ (assuming $\cos(\pi x) \neq 0$): $\tan(\pi x) = -1$ This equation has solutions at: $\pi x = -\frac{3\pi}{4}, \frac{\pi}{4} \Rightarrow x = -\frac{3}{4}, \frac{1}{4}$

So, the critical points are $x = -\frac{3}{4}$ and $x = \frac{1}{4}$.

Determine the sign of $f'(x)$ in intervals around these points:

1. For $x \in \left[-1, -\frac{3}{4}\right]$
2. For $x \in \left(-\frac{3}{4}, \frac{1}{4}\right]$
3. For $x \in \left(\frac{1}{4}, 1\right]$

Step 2: Second Derivative for Concavity

To determine concavity, compute the second derivative, $f''(x)$.

$f''(x) = -\pi^2 \sin(\pi x) + \pi^2 \cos(\pi x)$ $f''(x) = \pi^2 (\cos(\pi x) - \sin(\pi x))$

Set $f''(x) = 0$ to find potential inflection points:

$\cos(\pi x) - \sin(\pi x) = 0$ or $\cos(\pi x) = \sin(\pi x)$

This occurs when: $\tan(\pi x) = 1$ Solutions for this are: $\pi x = \frac{\pi}{4}, \frac{5\pi}{4} \Rightarrow x = \frac{1}{4}, \frac{5}{4}$

Since $x = \frac{5}{4}$ is outside $[-1, 1]$, only $x = \frac{1}{4}$ is relevant.

Solutions Summary:

a. Intervals of Increase/Decrease

• Decreasing: Intervals where $f'(x) < 0$.
• Increasing: Intervals where $f'(x) > 0$.

b. Local Minima and Maxima

• Local Minimum: $x$ value(s) where $f'(x)$ changes from negative to positive.
• Local Maximum: $x$ value(s) where $f'(x)$ changes from positive to negative.

c. Concavity Intervals

• Concave Down: Intervals where $f''(x) < 0$.
• Concave Up: Intervals where $f''(x) > 0$.

d. Inflection Points

• Inflection Points: Values of $x$ where $f''(x) = 0$ and concavity changes.

Would you like further details on calculating each specific interval or any specific steps?

Additional Practice Questions

1. How would you determine if a critical point is a maximum, minimum, or saddle point?
2. What is the significance of the second derivative in determining concavity?
3. How do we confirm if a point is an inflection point?
4. Can you explain why we solve $f'(x) = 0$ to find critical points?
5. How would you handle a problem if the interval was unbounded?

Tip:

Always analyze the sign of the derivative around critical points to confirm intervals of increase and decrease.