Problem 46 - Single Variable Calculus

Question

Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. $$ f(x)=\sin ^{2} x, \quad 0

Step-by-Step Solution

Answer

The function \(f(x) = \sin^2x\) has a relative maximum at \(x = \frac{\pi}{2}\) and a relative minimum at \(x = \pi\) in the interval \((0, \frac{3\pi}{2})\).

Step 1: Find the first derivative of the function

Differentiate \(f(x) = \sin^2x\) with respect to x to find the first derivative, \(f'(x)\). Applying the chain rule, we get: \(f'(x) = 2 \sin x \cdot \cos x\)

Step 2: Find the critical points

Set \(f'(x) = 0\) and solve for x. This will yield the critical points of the function. \(2 \sin x \cos x = 0\) The equation is satisfied when either \(\sin x = 0\) or \(\cos x = 0\). In the given interval \((0, \frac{3\pi}{2})\), the possible solutions are: \[\sin x = 0 \Rightarrow x = \pi\] \[\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}\] So, we have three critical points: \(x = \pi, \frac{\pi}{2}, \frac{3\pi}{2}\). However, since our interval is open (\(0 < x < \frac{3\pi}{2}\)), only the critical points \(x = \pi\) and \(x = \frac{\pi}{2}\) are valid.

Step 3: Find the second derivative of the function

Differentiate \(f'(x) = 2 \sin x \cos x\) with respect to x to find the second derivative, \(f''(x)\). Applying the product rule, we get: \(f''(x) = (2\cos x)\cos x + (2 \sin x)(-\sin x)\) \(f''(x) = 2\cos^2x - 2\sin^2x\)

Step 4: Determine the concavity at critical points

We'll plug the critical points \(x = \pi\) and \(x = \frac{\pi}{2}\) into the second derivative. For \(x = \pi\): \(f''(\pi) = 2\cos^2(\pi) - 2\sin^2(\pi) = 2\) Since \(f''(\pi) > 0\), the function is concave up at \(x = \pi\). Therefore, there is a relative minimum at \(x = \pi\). For \(x = \frac{\pi}{2}\): \(f''(\frac{\pi}{2}) = 2\cos^2(\frac{\pi}{2}) - 2\sin^2(\frac{\pi}{2}) = -2\) Since \(f''(\frac{\pi}{2}) < 0\), the function is concave down at \(x = \frac{\pi}{2}\). Therefore, there is a relative maximum at \(x = \frac{\pi}{2}\).

Step 5: Conclusion

The function \(f(x) = \sin^2x\) has a relative maximum at \(x = \frac{\pi}{2}\) and a relative minimum at \(x = \pi\) in the interval \((0, \frac{3\pi}{2})\).

Key Concepts

First Derivative

Understanding the concept of the first derivative is crucial when analyzing the behavior of functions. The first derivative of a function, denoted as \( f'(x) \), represents the rate at which the function's value changes with respect to a change in the independent variable, usually \( x \). Simply put, it measures how fast or slow the function's output (y-value) changes as you move along the curve.

For the exercise in question, the first derivative of the function \( f(x) = \text{sin}^2(x) \) is found to be \( f'(x) = 2 \text{sin}(x) \text{cos}(x) \). This step is foundational because the first derivative is essential for finding critical points where the function's slope is zero or undefined - these points are possible candidates for relative extrema, where the function could be at a high or low point in the interval of interest.

Critical Points

Critical points are the x-values where the first derivative of a function is either zero or undefined. These points are significant because they are the potential locations of the function's local maximums, minimums, or inflection points.

In our example, setting the first derivative \( f'(x) \) equal to zero leads us to the critical points. The equation \( 2 \text{sin}(x) \text{cos}(x) = 0 \) gives us the critical points within the interval \( 0 < x < \frac{3\text{\pi}}{2} \), specifically at \( x = \text{\pi} \) and \( x = \frac{\text{\pi}}{2} \). It is always important to remember the given interval, as any critical points found outside of this interval are not considered when evaluating relative extrema.

Second Derivative Test

The Second Derivative Test is a method used to determine whether a function's critical points are relative maximums, minimums, or neither. The test involves examining the concavity of the function at the critical points by using the second derivative, \( f''(x) \).

For a critical point \( c \), if \( f''(c) > 0 \), then the function is concave up and has a relative minimum at \( c \). Conversely, if \( f''(c) < 0 \), the function is concave down, indicating a relative maximum at \( c \). If the second derivative is zero, the test is inconclusive, and we may need to use other methods to analyze the function's behavior at \( c \). In the given exercise, the second derivative test correctly identifies a relative minimum at \( x = \text{\pi} \) and a relative maximum at \( x = \frac{\text{\pi}}{2} \), thereby proving itself as an effective tool for analyzing relative extrema.

Concavity

Concavity refers to the curvature direction of the graph of a function. If the graph of a function curves upward, like a smile, it is said to be concave up. If it curves downward, like a frown, it is concave down. A function that is concave up will hold water like a cup, while one that is concave down will not.

For our exercise, we determine the concavity by evaluating the second derivative at critical points. The sign of the second derivative tells us the concavity: if \( f''(x) > 0 \), the function is concave up at \( x \), and if \( f''(x) < 0 \), it is concave down. This understanding of concavity aids in classifying the relative extrema found at critical points.