Problem 231 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. \(f(x)=\sin (\pi x)-\cos (\pi x)\) over \(x=[-1,1]\)

Step-by-Step Solution

Answer

The function is always increasing. Inflection points are at \(x = -0.25\) and \(x = 0.75\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we start by finding the first derivative of the function. Given the function: \[ f(x) = \sin(\pi x) - \cos(\pi x) \]The first derivative is: \[ f'(x) = \pi \cos(\pi x) + \pi \sin(\pi x) \] This simplifies to: \[ f'(x) = \pi(\cos(\pi x) + \sin(\pi x)) \]

Step 2: Determine Critical Points

To find critical points, set the derivative equal to zero:\[ f'(x) = \pi(\cos(\pi x) + \sin(\pi x)) = 0 \]Thus,\[ \cos(\pi x) + \sin(\pi x) = 0 \]Solving the equation,\[ \tan(\pi x) = -1 \]The solutions in the interval \([-1, 1]\) are: \[ x = -0.75, \, x = 0.25 \]

Step 3: Test Intervals Around Critical Points

Test intervals around the critical points to determine whether the function is increasing or decreasing:- For \(x\) in \([-1, -0.75)\), pick \(x = -0.9\):\[ f'(-0.9) = \pi(\cos(-0.9\pi) + \sin(-0.9\pi)) > 0 \] (function increasing)- For \(x\) in \((-0.75, 0.25)\), pick \(x = 0\):\[ f'(0) = \pi(1) = \pi > 0 \] (function increasing)- For \(x\) in \((0.25, 1]\), pick \(x = 0.5\):\[ f'(0.5) = \pi(\cos(0.5\pi) + \sin(0.5\pi)) = 0 \] No sign change; still increasing.

Step 4: Identify and Analyze Extrema

Since the function is increasing across the entire domain, there are no local maxima or minima within \([-1, 1]\).

Step 5: Find the Second Derivative

To examine concavity, calculate the second derivative:\[ f''(x) = -\pi^2(\sin(\pi x) - \cos(\pi x)) \]

Step 6: Determine Concavity and Points of Inflection

Set the second derivative equal to zero to find potential inflection points:\[ -\pi^2(\sin(\pi x) - \cos(\pi x)) = 0 \] Simplifying, we get:\[ \sin(\pi x) = \cos(\pi x) \]Solving,\[ \tan(\pi x) = 1 \]The solutions in the interval \([-1, 1]\) are \[ x = -0.25, \, x = 0.75 \]Test intervals around these points to determine concavity:- For \(x\) in \([-1, -0.25)\), select \(x = -0.5\); \(f''(-0.5) < 0\) => concave down.- For \(x\) in \((-0.25, 0.75)\), select \(x = 0\); \(f''(0) > 0\) => concave up.- For \(x\) in \((0.75, 1]\), select \(x = 0.9\); \(f''(0.9) < 0\) => concave down.

Key Concepts

First Derivative

The first derivative of a function is a powerful tool in calculus. It not only tells us the rate of change of the function but also reveals where the function is increasing or decreasing. For our function, \( f(x) = \sin(\pi x) - \cos(\pi x) \), the first derivative is found using basic derivative rules for sine and cosine:

• \( f'(x) = \pi \cos(\pi x) + \pi \sin(\pi x) \)

This expression simplifies to \( f'(x) = \pi(\cos(\pi x) + \sin(\pi x)) \). The first derivative helps us analyze the function more deeply:

• We can set \( f'(x) = 0 \) to find critical points.
• The sign of \( f'(x) \) tells us whether the function is increasing (when \( f'(x) > 0 \)) or decreasing (when \( f'(x) < 0 \)).

Understanding the behavior of the first derivative provides insights into the function's graph and character.

Critical Points

Critical points of a function are found where its derivative equals zero or does not exist. These points often correspond to local minima or maxima, or neither, depending on other criteria. For the function given, critical points occur when:

• \( \pi(\cos(\pi x) + \sin(\pi x)) = 0 \)
• Solving leads to \( \tan(\pi x) = -1 \)

Within the interval \([-1, 1]\), the critical points are \( x = -0.75 \) and \( x = 0.25 \). To determine whether these points are local extrema, we look at the intervals around these points to check the sign change of the derivative:

• If \( f'(x) \) changes from positive to negative, it indicates a local maximum.
• If \( f'(x) \) changes from negative to positive, it indicates a local minimum.
• If there is no sign change, the point is neither a local maxima nor minima.

This critical point analysis helps sketch the function and understand its localized behavior.

Concavity

Concavity focuses on how a function curves, indicating where it is "open up" (concave up) or "open down" (concave down). The second derivative is the tool for this analysis, as it reveals the nature of concavity:

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

Evaluating this second derivative, certain rules apply:

• \( f''(x) > 0 \) implies that the function is concave up in that interval.
• \( f''(x) < 0 \) implies concave down in that interval.

By examining intervals around potential inflection points, you can decide how the function behaves across its domain.

Inflection Points

Inflection points are where the graph of the function changes concavity. To find such points, set the second derivative equal to zero and solve for \( x \):

• \( -\pi^2(\sin(\pi x) - \cos(\pi x)) = 0 \)
• Simplifies to \( \sin(\pi x) = \cos(\pi x) \)
• Giving \( \tan(\pi x) = 1 \)

Within the range \([-1, 1]\), the solutions are \( x = -0.25 \) and \( x = 0.75 \). At each of these points, the concavity changes, effectively marking them as points of inflection. Checking the sign of the second derivative on intervals between these points confirms the shift in concavity:

• If \( f''(x) \) goes from positive to negative, this denotes a change from concave up to concave down.
• Conversely, if \( f''(x) \) goes from negative to positive, it signifies a change from concave down to concave up.

Understanding inflection points is critical in capturing the complete picture of a function's graphical behavior.