Problem 53 - Essential Calculus

Question

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, (b) 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 \(f(x)=\sin^{2}x+\sin x\) on the interval \((0,2\pi)\) increases on \((0, \frac{\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\), and decreases on \((\frac{\pi}{2}, \frac{3\pi}{2})\). The relative maximum occurs at \(\frac{\pi}{2}\) and relative minimum occurs at \(\frac{3\pi}{2}\).

Step 1: Differentiation

The first step is to find the derivative of the function \(f(x)=\sin ^{2} x+\sin x\). The derivative of \(\sin x\) is \(\cos x\), and using the chain rule, the derivative of \(\sin^2 x\) is \(2\sin x\cos x\). With this, we get \(f'(x)=2\sin x \cos x + \cos x\).

Step 2: Finding critical points

Next, we find the critical points by setting the first derivative equal to zero: \(0 = 2\sin x \cos x + \cos x\), which simplifies to \(0 = \cos x(2 \sin x + 1)\). From this equation, critical points occur when \(\cos x = 0\) or \((2\sin x + 1) = 0 \), which gives us \(x = \frac{\pi}{2}, \frac{3\pi}{2}, –\frac{1}{2}\). Since \(-\frac{1}{2}\) doesn't fall within the interval (0,2\pi), our critical points are \(x = \frac{\pi}{2}\), and \(\frac{3\pi}{2}\).

Step 3: First Derivative Test

Now we determine the intervals of increase and decrease. This can be done by taking values in the intervals \((0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi)\) and evaluating them in the derivative function. After doing so we find that the function increases on the interval \((0, \frac{\pi}{2})\), decreases on \((\frac{\pi}{2}, \frac{3\pi}{2})\), and increases on \((\frac{3\pi}{2}, 2\pi)\). The relative maximum occurs at \((\frac{\pi}{2}, f(\frac{\pi}{2}))\) and relative minimum occurs at \((\frac{3\pi}{2}, f(\frac{3\pi}{2}))\)

Step 4: Confirm with Graph

Finally, we use a graphing utility to sketch the function within the given interval. The function’s critical points and areas of increase and decrease should align with the calculations.

Key Concepts

Critical Points

Critical points in calculus are specific values of the independent variable where the first derivative of a function is either zero or undefined. These points are essential because they often correspond to local maxima, local minima, or saddle points of the function. For a function, finding these points gives us valuable insights into its behavior.

To find the critical points of a function, you need to take the derivative and set it equal to zero. This helps identify where the rate of change, or slope, is zero, which is indicative of potential turning points. In some cases, these are the points where the function switches from increasing to decreasing, or vice versa.

• For instance, in the given function, known as the trigonometric function, after differentiating, we solved the equation his equation gives critical points at specific intervals.
• In the interval (0, 2\pi), we found critical points at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).

Understanding critical points provides a foundation for determining other key concepts, such as where a function is increasing or decreasing.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing involves analyzing the sign of its first derivative. As you calculate the derivative and find intervals where it is positive or negative, you can establish where the function exhibits various behaviors.

• If the first derivative \(f'(x) > 0\) in an interval, the function \(f(x)\) is increasing on that interval.
• Conversely, if \(f'(x) < 0\), the function is decreasing there.

After applying the First Derivative Test to the exercise function, we understood the intervals as:

• The function increases in the intervals \((0, \frac{\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\).
• It decreases in the interval \((\frac{\pi}{2}, \frac{3\pi}{2})\).

These distinctions become important when graphing the function or attempting to predict output for particular input ranges. Recognizing these intervals provides a clearer picture of the behavior and nature of the function across its domain.

Trigonometric Functions

Trigonometric functions are essential mathematical functions that relate the angles of a triangle to its sides. They are periodic, meaning their values repeat at regular intervals, which makes them particularly interesting for analysis like the one in the provided exercise.

• The sine function, \(\sin x\), one of the key trigonometric functions, oscillates between -1 and 1.
• Functions like \(\sin^{2}x + \sin x\) combine these trigonometric elements to create new periodic behaviors.

When dealing with such functions in calculus, certain steps help to reveal their unique features, such as finding where they increase and decrease within a given domain. Understanding these properties is crucial for mathematicians, scientists, and engineers, who frequently apply these principles in fields spanning physics, signal processing, and other areas that rely on periodicity and oscillation. Remember that with trigonometric functions, numerous behavior activities can be logically predicted by considering their periodic nature and influences from combined trigonometric elements.