Problem 57 - University Calculus: Early Transcendentals

Question

a. Find the local extrema of each function on the given interval, and say where they occur. b. Graph the function and its derivative together. Comment on the behavior of \(f\) in relation to the signs and values of \(f^{\prime}\) $$f(x)=\sin 2 x, \quad 0 \leq x \leq \pi$$

Step-by-Step Solution

Answer

Local maximum at \( x = \frac{\pi}{4} \) and local minimum at \( x = \frac{3\pi}{4} \).

Step 1: Find the Derivative

First, we need the derivative of \( f(x) = \sin 2x \). Using the chain rule, the derivative of \( \sin u \) where \( u = 2x \) is \( \cos u \cdot \frac{du}{dx} \). Thus, \( f'(x) = 2\cos 2x \).

Step 2: Set Derivative Equal to Zero

To find critical points, set the derivative \( f'(x) = 2\cos 2x = 0 \). Solving for \( x \) yields \( \cos 2x = 0 \). This occurs when \( 2x = \frac{\pi}{2} + n\pi \) for integers \( n \). Thus, \( x = \frac{\pi}{4} + \frac{n\pi}{2} \).

Step 3: Solve for Critical Points Within Interval

Restrict the solutions from Step 2 to the interval \([0, \pi]\). The values satisfying \( x = \frac{\pi}{4}, \frac{3\pi}{4} \) are within the interval. These are the critical points.

Step 4: Evaluate Function at Critical Points

Evaluate \( f(x) = \sin 2x \) at \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \) to find local extrema. \( f(\frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1 \) and \( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{2}) = -1 \).

Step 5: Test Endpoints

Evaluate \( f(x) \) at the endpoints: \( x = 0 \) and \( x = \pi \). \( f(0) = \sin(0) = 0 \) and \( f(\pi) = \sin(2\pi) = 0 \).

Step 6: Determine Local Extrema

With evaluations: \( f(\frac{\pi}{4}) = 1 \) is a local maximum, and \( f(\frac{3\pi}{4}) = -1 \) is a local minimum. The function has zeros at endpoints.

Step 7: Graph Function and Derivative

Graph \( f(x) = \sin 2x \) and its derivative \( f'(x) = 2\cos 2x \). Notice that when \( f(x) \) is increasing, \( f'(x) > 0 \), and when \( f(x) \) is decreasing, \( f'(x) < 0 \). At the points where \( f'(x) = 0 \), \( f(x) \) has local extrema.

Key Concepts

Critical Points

Critical points are where a function's derivative equals zero or where the derivative does not exist. These points are essential for finding local extrema, which refer to the maximum or minimum values a function attains at specific points within a given interval.
To find the critical points, you start by taking the derivative of the function. For the function \( f(x) = \sin 2x \), we calculated the derivative \( f'(x) = 2\cos 2x \) using the chain rule. We then set the derivative equal to zero: \( 2\cos 2x = 0 \).
Solving for \( x \), we found \( x = \frac{\pi}{4} + \frac{n\pi}{2} \) (where \( n \) is an integer), which are potential critical points.
Restricting it to the interval \( [0, \pi] \) gives the critical points \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \). These are crucial in determining the local extrema of the function.

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function changes at any point. It’s essential in identifying the behavior of a function, especially when looking for local extrema.
In our example, the derivative \( f'(x) = 2\cos 2x \) tells us how \( f(x) = \sin 2x \) changes with respect to \( x \). When \( f'(x) > 0 \), the function \( f(x) \) is increasing, and when \( f'(x) < 0 \), the function is decreasing.
This changing behavior helps us pinpoint the critical points, which occur when \( f'(x) = 0 \). These points are potential places where the function's slope is zero, indicating possible maxima, minima, or points of inflection. The derivative's sign thus gives us a comprehensive understanding of the function's overall behavior across the interval.

Chain Rule

The chain rule is an essential tool in calculus used to find the derivative of composite functions. It states that if you have a function \( g(x) \) nested inside another function \( f(x) \), the derivative of this composite function \( f(g(x)) \) is found by taking the derivative of \( f \) with respect to \( g(x) \), and then multiplying it by the derivative of \( g(x) \).
In our problem, we applied the chain rule to differentiate \( f(x) = \sin(2x) \). Here, \( f(u) = \sin u \) and \( u = 2x \).
Following the chain rule, the derivative \( f'(x) \) becomes \( \frac{d}{du} \sin(u) \cdot \frac{du}{dx} \). Simplifying, we get \( \cos u \cdot 2 = 2\cos(2x) \).
This expression gives us the derivative needed to locate the critical points and understand the behavior of the original function.

Interval Analysis

Interval analysis involves accurately examining how a function behaves over a specific range. It's a vital step in finding local extrema by ensuring calculations and findings fit within the defined boundaries of an interval.
After finding critical points for \( f(x) = \sin 2x \), we must verify which points fall within the given interval \([0, \pi]\).
In this analysis, only the critical points \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \) lie within the specified interval. Additionally, we also evaluate the function at the interval's endpoints, \( x = 0 \) and \( x = \pi \).
This process helps identify not only local extrema but also how the endpoints contribute to the function's behavior. Interval analysis ensures a comprehensive examination of the function's ups and downs within its entire range, guiding us to correctly assess its local maxima and minima.