Problem 47 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(41-48\) , consider the function on the interval (0,2 \pi). For each function, (a) find the open interval(s) on which the function is increasing or decreasing, apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results. $$ f(x)=\sin ^{2} x+\sin x $$

Step-by-Step Solution

Answer

The function \( f(x) = \sin^2x + \sin x \) is increasing on the intervals (0, \( \pi/2 \)) and (\( \pi, 3\pi/2 \)), and decreasing on the intervals (\( \pi/2, \pi \)) and (\( 3\pi/2, 2\pi \)). It has relative maximum at \(x=\pi/2\) and relative minima at \(x=\pi\) and \(x = 2\pi \).

Step 1: Compute the Derivative

One starts by obtaining the derivative of the function \( f(x) = \sin^2x + \sin x \). Using the power rule and chain rule for differentiation, the derivative \( f'(x) \) is found to be \( f'(x) = 2\sin x \cos x + \cos x = \cos x (2\sin x + 1) \).

Step 2: Identify Critical Points

Critical points are found by setting the derivative \( f'(x) =0 \). Solving the equation \( \cos x (2\sin x + 1) = 0 \), we obtain critical points at \(x = \pi /2, 3\pi /2, \pi \) in the interval (0, \( 2\pi \)).

Step 3: Determine the Sign of the Derivative

One needs to determine whether the derivative is positive or negative. From \(x = 0\) to \(x = \pi/2, \cos x > 0\) and \(2\sin x + 1 > 0\), so \(f'(x) > 0\), indicating that \(f(x)\) is increasing. From \(x = \pi/2\) to \(x = \pi, \cos x > 0\) and \(2\sin x + 1 < 0\), thus \(f'(x) < 0\) and \(f(x)\) is decreasing. From \(x = \pi\) to \(x = 3\pi/2, \cos x < 0\) and \(2\sin x + 1 < 0\), so \(f'(x) > 0\), implying that \(f(x)\) is increasing. From \(x = 3\pi/2\) to \(x = 2\pi, \cos x < 0\) and \(2\sin x + 1 > 0\), thus \(f'(x) < 0\), showing that \(f(x)\) is decreasing.

Step 4: Identify the Extrema

Once knowing on which intervals the function is increasing or decreasing, one can identify the local maxima and minima using the First Derivative Test. \(f(x)\) has a relative maximum at \(x=\pi/2\) and a relative minimum at \(x=\pi\) and \(x = 2\pi \).

Step 5: Verify with a Graphing Utility

Finally, these results should be verified using a graphing utility. The graph should show a relative maximum at \(x=\pi/2\) and relative minima at \(x=\pi\) and \(x = 2\pi \). The locations of increasing and decreasing functions should also correspond to those predicted.

Key Concepts

Increasing and Decreasing Functions

When studying functions, it is essential to understand when a function is increasing or decreasing over certain intervals. A function is said to be increasing on an interval if the output of the function becomes larger as the input increases. Conversely, the function is decreasing when the output decreases as the input increases.

In calculus, this behavior of functions is determined by looking at the function's first derivative, represented by the symbol \(f'(x)\). If \(f'(x) > 0\) on an interval, the function is increasing on that interval. If \(f'(x) < 0\), the function is decreasing. The exercise provided involves this analysis for the function \(f(x) = \sin^2x + \sin x\), where the intervals of increase and decrease are established by examining the sign of the first derivative.

Critical Points in Calculus

Critical points play a significant role when analyzing the behavior of functions. In calculus, a critical point occurs where the derivative of a function is either zero or undefined. At these points, functions can change from increasing to decreasing or vice versa, and they are also potential locations for finding local maxima and minima, which are collectively known as relative extrema.

To find critical points, as seen in our exercise, set the first derivative equal to zero and solve for \(x\). For the function \(f(x)\), the derivative \(f'(x) = \cos x (2\sin x + 1)\) results in the critical points \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \pi\) within the interval \(0, 2\pi\). Understanding and identifying critical points allows us to determine where to apply the First Derivative Test.

Relative Extrema

The relative extrema of a function include the relative maxima and minima—points where the function reaches a local highest or lowest value, respectively. The First Derivative Test is a valuable tool to identify these points. After finding the critical points, use the test by examining the sign changes of \(f'(x)\) around each critical point.

For a relative maximum, \(f'(x)\) will change from positive to negative, and for a relative minimum, it will change from negative to positive as \(x\) passes through the critical value. In the given exercise, we find that \(f(x)\) has a relative maximum at \(x = \frac{\pi}{2}\) because \(f'(x)\) changes from positive to negative. Similarly, relative minima are found at \(x=\pi\) and \(x = 2\pi\). This analysis is crucial for understanding the overall shape of the graph of the function.

Derivative of Trigonometric Functions

The derivatives of trigonometric functions are foundational within calculus. For the function \(f(x) = \sin^2x + \sin x\), knowledge of these derivatives is necessary to perform the differentiation. When dealing with trigonometric functions, it is important to remember certain rules like the chain rule and product rule.

The derivative of \(\sin x\) is \(\cos x\), and using the chain rule, the derivative of \(\sin^2x\) results in \(2\sin x\cos x\). Thus, the derivative \(f'(x) = 2\sin x \cos x + \cos x\) is obtained, which allows for further analysis in calculus tasks such as identifying critical points and understanding the function's increasing and decreasing behavior as previously discussed.