Problem 27 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\sin x \cos x, \quad 0 \leq x \leq \pi$$

Step-by-Step Solution

Answer

Local extrema are at \(x=\frac{\pi}{4}\) (max) and \(x=\frac{3\pi}{4}\) (min). Inflection point at \(x=\frac{\pi}{2}\).

Step 1: Find the first derivative

To find the critical points of the function \(y = \sin x \cos x\), we first need to find its first derivative. Using the product rule \((uv)' = u'v + uv'\), where \(u = \sin x\) and \(v = \cos x\), we differentiate to obtain: \[y' = \cos^2 x - \sin^2 x\]This can be simplified using the identity \(\cos(2x) = \cos^2 x - \sin^2 x\), so the derivative is:\[y' = \cos(2x)\]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Set \(\cos(2x) = 0\) and solve for \(x\):\[2x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}\]Simplifying, we find:\[x = \frac{\pi}{4} + k\frac{\pi}{2}, \quad k \in \mathbb{Z}\]Within the interval \(0 \le x \le \pi\), the relevant critical points are \(x = \frac{\pi}{4}, \frac{3\pi}{4}\).

Step 3: Determine the nature of critical points

Apply the first derivative test or analyze using the function values:1. At \(x = 0\) and \(x = \pi\), \(y = 0\).2. At \(x = \frac{\pi}{4}\), \(y = \frac{1}{2}\). 3. At \(x = \frac{3\pi}{4}\), \(y = -\frac{1}{2}\).The value at \(x = \frac{\pi}{4}\) is a local maximum and \(x = \frac{3\pi}{4}\) is a local minimum, considering \(y(0) = y(\pi) = 0\), which also suggests an inflection-like transition.

Step 4: Find the second derivative

To determine inflection points, we find the second derivative.Differentiate the first derivative \(y' = \cos(2x)\):\[y'' = -2\sin(2x)\]Set \(-2\sin(2x) = 0\) to solve for inflection points within \(0 \le x \le \pi\).\[\sin(2x) = 0 \Rightarrow 2x = n\pi, \quad n \in \mathbb{Z} \Rightarrow x = \frac{n\pi}{2}, \quad n \in \{0, 1, 2\}\]The possible values here are \(x = 0, \frac{\pi}{2}, \pi\), where \(\sin x \cos x = 0\) for \(x = 0\) and \(x = \pi\). Hence, \(x = \frac{\pi}{2}\) acts like an inflection point as \(\sin(2x)\) changes sign across this point.

Step 5: Graph the function

Plot the function \(y = \sin x \cos x\) over the interval \(0 \le x \le \pi\). Mark the critical points: local maximum at \(x = \frac{\pi}{4}\) with \(y = \frac{1}{2}\), and local minimum at \(x = \frac{3\pi}{4}\) with \(y = -\frac{1}{2}\). Also note inflection-like behavior at \(x = \frac{\pi}{2}\), transitioning from concave down to concave up. The graph touches the x-axis at \(x = 0\) and \(x = \pi\), reflecting the fact that these boundary points coincide with zero values.

Key Concepts

Critical Points

In calculus, critical points of a function are where the derivative is either zero or undefined. These points are crucial to finding the local maximums or minimums, as they mark where the function switches between increasing and decreasing. For the function \(y = \sin x \cos x\), we began by finding the first derivative using the product rule, yielding \(y' = \cos(2x)\). To find the critical points, we solve \(\cos(2x) = 0\), which leads us to:

• \( x = \frac{\pi}{4} \)
• \( x = \frac{3\pi}{4} \)

These are the points in the interval \([0, \pi]\) where the slope of the tangent line is zero. Upon evaluating the function at these points, \(y\left(\frac{\pi}{4}\right) = \frac{1}{2}\) portrays a local maximum, and \(y\left(\frac{3\pi}{4}\right) = -\frac{1}{2}\) reflects a local minimum. This suggests a peak and a valley within the interval, highlighting the behavior changes of the function.

Inflection Points

Inflection points occur where the second derivative changes sign, indicating a shift in the concavity of the function. For the given function \(y = \sin x \cos x\), the second derivative is \(y'' = -2\sin(2x)\). To detect inflection points, we solve \(-2\sin(2x) = 0\). This simplifies to:

• \( x = 0 \)
• \( x = \frac{\pi}{2} \)
• \( x = \pi \)

Among these, \(x = \frac{\pi}{2}\) acts as an inflection point as the derivative \(\sin(2x)\) transitions through zero, changing its sign. This signifies a shift from concave down to concave up, representing where the graph of the function bends.

Graphing Functions

Graphing functions provides a visual representation of how they behave across intervals. For the function \(y = \sin x \cos x\), this aids in recognizing critical areas like maxima, minima, and inflection points. When plotting over the interval \([0, \pi]\), key features include:

• The local maximum at \(x = \frac{\pi}{4}\), where \(y = \frac{1}{2}\).
• The local minimum at \(x = \frac{3\pi}{4}\), where \(y = -\frac{1}{2}\).
• Inflection-like behavior at \(x = \frac{\pi}{2}\), where concavity shifts.
• Intersections with the x-axis at \(x = 0\) and \(x = \pi\), shown by the function equating to zero.

These elements help define the wave-like pattern of the graph, capturing variations in amplitude and inflection through the specified domain.