Problem 35 - Calculus A New Horizon

Question

Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$\sin x+\cos x$$

Step-by-Step Solution

Answer

The function has relative extrema at \( x = \frac{\pi}{4} + n\pi \) and inflection points at \( x = \frac{3\pi}{4} + n\pi \).

Step 1: Find the First Derivative

To find the critical points and determine relative extrema, we first need to find the first derivative of the function. The function given is \( f(x) = \sin x + \cos x \). The derivative of \( \sin x \) is \( \cos x \) and the derivative of \( \cos x \) is \( -\sin x \). Hence, the first derivative \( f'(x) \) is \( \cos x - \sin x \).

Step 2: Find Critical Points

To find critical points, we set the first derivative equal to zero: \( \cos x - \sin x = 0 \). This simplifies to \( \cos x = \sin x \). This equation is true for \( x = \frac{\pi}{4} + n\pi \), where \( n \) is an integer.

Step 3: Determine the Second Derivative

The second derivative helps identify concavity and inflection points. We find the second derivative by differentiating \( f'(x) = \cos x - \sin x \), giving us \( f''(x) = -\sin x - \cos x \).

Step 4: Find Inflection Points

Set the second derivative equal to zero to find inflection points: \( -\sin x - \cos x = 0 \), which simplifies to \( \sin x = -\cos x \). This is true for \( x = \frac{3\pi}{4} + n\pi \), where \( n \) is an integer.

Step 5: Analyze Critical and Inflection Points

Plugging \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \) into the second derivative, we determine the concavity: \( f''(\frac{\pi}{4}) = -\sqrt{2} \) is negative, indicating \( x = \frac{\pi}{4} \) is a relative maximum, and \( f''(\frac{3\pi}{4}) = 0 \) is an inflection point.

Step 6: Sketch the Graph

Using the relative extrema and inflection points \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \), sketch the graph of \( f(x) = \sin x + \cos x \). The graph oscillates with these points determining its peaks and changes in concavity.

Step 7: Verify with a Graphing Utility

Use a graphing tool to verify the locations of critical points, relative extrema, and inflection points. The tool should show peaks around \( x = \frac{\pi}{4} + n\pi \) and inflection around \( x = \frac{3\pi}{4} + n\pi \).

Key Concepts

Derivatives

Derivatives form the bedrock of calculus and are crucial in understanding how functions change. Imagine a function as a road; the derivative tells us about the slope of that road at any point.
For our function \( f(x) = \sin x + \cos x \), we calculate its first derivative to see where the slope might be zero—indicating possible peaks or valleys.

• The derivative \( f'(x) = \cos x - \sin x \) shows us the rate of change of our function.
• To find critical points, set \( f'(x) = 0 \), leading to \( \cos x = \sin x \).
• This equation is satisfied at points \( x = \frac{\pi}{4} + n\pi \), where \( n \) is an integer.

By identifying these critical points, we can further explore the function’s graph and behavior.

Inflection Points

Inflection points are segments where the curve changes its concavity—like how a smile flips into a frown. To find these points, we need the second derivative of the function.
By differentiating our first derivative \( f'(x) = \cos x - \sin x \), we get the second derivative \( f''(x) = -\sin x - \cos x \).

• Setting \( f''(x) = 0 \) helps find inflection points: \( \sin x = -\cos x \).
• Solutions occur at \( x = \frac{3\pi}{4} + n\pi \), where \( n \) is an integer.

Inflection points mark the spot where the graph's concavity changes; thus, they are key in sketching the graph's overall shape.

Relative Extrema

Relative extrema refer to the peaks and valleys of a function—points where the function reaches a local maximum or minimum. To find relative extrema, first identify critical points using the first derivative.
For \( f(x) = \sin x + \cos x \), critical points occur at \( x = \frac{\pi}{4} + n\pi \). At these points, we further examine the second derivative:

• If \( f''(x) \) is negative, the critical point is a relative maximum.
• If \( f''(x) \) is positive, it indicates a relative minimum.

In our case, \( f''(\frac{\pi}{4}) = -\sqrt{2} \) confirms a relative maximum, giving us insight into the peaks of the graph.

Graphing Functions

Graphing functions provides a visual representation of mathematical behavior. It can be seen as painting a picture based on everything learned from derivatives and inflection points.
For our function, pinpointing the location of relative maxima and inflection points serves as guides to sketch its shape.

• Relative maxima around \( x = \frac{\pi}{4} + n\pi \) create peaks.
• Inflection points at \( x = \frac{3\pi}{4} + n\pi \) hint at changes in the graph's curvature.

Using graphing utilities aligns our sketch with the actual oscillating wave pattern of the function, ensuring accuracy and completeness in representing its behavior.