Problem 12 - Calculus Single Variable

Question

Follow the outline given in this section to give a careful sketch of the graph of each of the functions in Exercises \(1-24\). Your sketch should exhibit, and have labeled, all of the following: a) local and global extrema, b) inflection points, c) intervals on which function is increasing or decreasing, d) intervals on which function is concave up or concave down, e) horizontal and vertical asymptotes. $$ f(x)=\cos (x)-\sin (x) $$

Step-by-Step Solution

Answer

The function has local minima at \( x = \frac{3\pi}{4} + n\pi \), inflection points at \( x = \frac{\pi}{4} + n\pi \), increases and decreases periodically, is concave up and down between these points, with no asymptotes.

Step 1: Find Critical Points

To find the extrema and intervals on which the function is increasing or decreasing, first find the derivative of the function: \[ f'(x) = -\sin(x) - \cos(x). \] Find the critical points by setting the derivative equal to zero: \[ -\sin(x) - \cos(x) = 0 \] or \[ \sin(x) + \cos(x) = 0. \] Solving this equation gives critical points at \( x = \frac{3\pi}{4} + n\pi \), where \( n \) is an integer.

Step 2: Determine Extrema

To determine whether the critical points are local maxima, minima, or neither, use the second derivative test: \[ f''(x) = -\cos(x) + \sin(x). \] Evaluate \( f''(x) \) at the critical points. For \( x = \frac{3\pi}{4} \), \( f''(\frac{3\pi}{4}) > 0 \), which indicates a local minimum at these points. Repeat this for other critical points to find their nature.

Step 3: Identify Inflection Points

An inflection point occurs where the concavity changes, found by setting the second derivative to zero: \[ -\cos(x) + \sin(x) = 0 \] or \[ \sin(x) = \cos(x). \] Solving this gives \( x = \frac{\pi}{4} + n\pi \) (where \( n \) is an integer) for possible inflection points. Check intervals around these points to confirm changes in concavity.

Step 4: Determine Intervals of Increase/Decrease

Using the derivative found in Step 1, analyze intervals between the critical points. For example, when testing intervals like \((0, \frac{\pi}{4})\) and \( (\frac{\pi}{4}, \frac{3\pi}{4}) \), find that the function decreases in intervals where \( f'(x) < 0 \) and increases where \( f'(x) > 0 \).

Step 5: Define Concavity Intervals

Use the second derivative found in Step 3 to determine the concavity. In the intervals where \( f''(x) > 0 \), the function is concave up, and where \( f''(x) < 0 \), the function is concave down. Use test points in each interval to determine concavity.

Step 6: Locate Asymptotes

Examine the function for asymptotes. As \( f(x) = \cos(x) - \sin(x) \) is a continuous trigonometric function, it has no vertical or horizontal asymptotes.

Step 7: Sketch the Graph

With all the above information, sketch the function by marking the critical points, inflection points, intervals of increase/decrease, and areas of concavity. Label all features accordingly.

Key Concepts

Critical Points

In calculus, critical points are where the derivative of a function is zero or undefined. For the function \( f(x) = \cos(x) - \sin(x) \), we determine the critical points by taking its derivative \( f'(x) = -\sin(x) - \cos(x) \) and setting it to zero: \[ -\sin(x) - \cos(x) = 0. \]This simplifies to \( \sin(x) + \cos(x) = 0 \). Solving this equation gives us critical points at \( x = \frac{3\pi}{4} + n\pi \), where \( n \) is any integer. At these points, the rate of change of the function is zero, meaning the function does not increase or decrease momentarily. These critical points help identify local peaks and valleys, crucial for sketching the graph.

Derivative Analysis

Derivative analysis involves understanding what the first derivative reveals about a function. It shows where the function is increasing or decreasing. For the function \( f(x) = \cos(x) - \sin(x) \), the derivative we calculated is \( f'(x) = -\sin(x) - \cos(x) \).The derivative being positive means the function is increasing; negative indicates it is decreasing. To analyze the intervals of increase and decrease, we test values between critical points. Consider intervals like \((0, \frac{\pi}{4})\) and \((\frac{\pi}{4}, \frac{3\pi}{4})\). By plugging test points into \( f'(x) \), we can determine:

• When \( f'(x) > 0 \) in an interval, the function is increasing.
• When \( f'(x) < 0 \), it is decreasing.

This information is instrumental in visualizing the graph's slope between critical points.

Inflection Points

Inflection points occur where the graph of a function changes concavity, i.e., from concave up to concave down or vice versa. To find inflection points for \( f(x) = \cos(x) - \sin(x) \), we look at the second derivative, \( f''(x) = -\cos(x) + \sin(x) \).Setting the second derivative to zero gives \[ -\cos(x) + \sin(x) = 0 \] or \[ \sin(x) = \cos(x). \] Solving this equation, we find potential inflection points at \( x = \frac{\pi}{4} + n\pi \), where \( n \) is an integer. To confirm actual inflection points, examine the intervals around these values. This helps determine where the function's curve changes from arching to dipping, a key feature in graph sketching.

Concavity Intervals

Concavity refers to the direction a graph curves and is determined by the second derivative. For \( f(x) = \cos(x) - \sin(x) \), the second derivative is \( f''(x) = -\cos(x) + \sin(x) \).To determine where the function is concave up or down:

• If \( f''(x) > 0 \) over an interval, the function is concave up (like the upward arch of a bowl).
• If \( f''(x) < 0 \), it is concave down (like an upside-down bowl).

Test points in intervals between \( x = \frac{\pi}{4} + n\pi \) (potential inflection points) By analyzing these intervals, we map out where the function curves upward or downward. This is crucial in creating an accurate sketch of the function's overall shape.