Problem 20 - 5 Steps to A 5: AP Calculus AB 2021

Question

Sketch the graphs of the following functions indicating any relative extrema, points of inflection, asymptotes, and intervals where the function is increasing, decreasing, concave upward, or concave downward. $$ f(x)=\cos x \sin ^{2} x[0,2 \pi] $$

Step-by-Step Solution

Answer

Question: Sketch the function $$f(x) = \cos{x} \sin^2{x}$$ on the interval [0, 2π], and describe the important features of the graph. Answer: The graph of $$f(x) = \cos{x} \sin^2{x}$$ on the interval [0, 2π] has the following important features: 1. Local maximum at x = π/2; f(π/2) = 0 2. Local minimum at x = 3π/2; f(3π/2) = 0 3. Points of inflection at x = 0, π/3, 2π/3, π, 4π/3, 5π/3, and 2π. 4. Increasing on the intervals (0, π/2) and (π, 3π/2) 5. Decreasing on the intervals (π/2, π) and (3π/2, 2π) 6. Concave upward in the ranges (0, π/3), (2π/3, π), and (π, 4π/3), (5π/3, 2π) 7. Concave downward in the ranges (π/3, 2π/3) and (4π/3, 5π/3) 8. No asymptotes are present in the graph.

Step 1: Calculate the first and second derivatives

First, find the first derivative of the function with respect to x: $$ f'(x) = \frac{d}{dx} (\cos{x} \sin^2{x}) = (-\sin{x}\sin^2{x} + 2\cos^2{x}\sin{x}) $$ Now, find the second derivative of the function with respect to x: $$ f''(x) = \frac{d^2}{dx^2} (\cos{x} \sin^2{x}) = -2\sin{x}\cos{x}(3\cos^2{x} - 1) $$

Step 2: Find the critical points

To find the critical points, set the first derivative equal to zero and solve for x: $$ f'(x) = 0 \implies -\sin{x}\sin^2{x} + 2\cos^2{x}\sin{x} = 0 $$ This equation becomes: $$ \sin{x}(3\cos^2{x} - \sin^2{x}) = 0 $$ We can factor this into: $$ \sin{x}(1-\sin^2{x})(3 - 3\sin^2{x}) = 0 $$ So, the critical points are at: x = 0, π/2, π, 3π/2, and 2π.

Step 3: Determine the local extrema

Now, utilize the first derivative test to determine if each critical point is a local maximum, local minimum or neither: - In the range (0, π/2), f'(x) > 0 implies f(x) is increasing. - In the range (π/2, π), f'(x) < 0 implies f(x) is decreasing. - In the range (π, 3π/2), f'(x) > 0 implies f(x) is increasing. - In the range (3π/2, 2π), f'(x) < 0 implies f(x) is decreasing. From these results, we can identify the local extrema: - At x = π/2; f(π/2) = 0 (Local maximum) - At x = 3π/2; f(3π/2) = 0 (Local minimum)

Step 4: Determine intervals where f(x) is increasing or decreasing

We already determined the intervals where the function is increasing and decreasing using the first derivative. So, - f(x) is increasing on the intervals: (0, π/2) and (π, 3π/2). - f(x) is decreasing on the intervals: (π/2, π) and (3π/2, 2π).

Step 5: Determine the points of inflection and concavity

Analyze the second derivative to find the points of inflection, where the concavity of the function changes: $$ f''(x) = 0 \implies -2\sin{x}\cos{x}(3\cos^2{x} - 1) = 0 $$ The points of inflection occur at x = 0, π/3, 2π/3, π, 4π/3, 5π/3, and 2π. Now, determine the intervals where f''(x) > 0 (concave upward) and where f''(x) < 0 (concave downward). - In the range (0, π/3), f''(x) > 0 implies f(x) is concave upward. - In the range (π/3, 2π/3), f''(x) < 0 implies f(x) is concave downward. - In the range (2π/3, π), f''(x) > 0 implies f(x) is concave upward. - In the range (π, 4π/3), f''(x) > 0 implies f(x) is concave upward. - In the range (4π/3, 5π/3), f''(x) < 0 implies f(x) is concave downward. - In the range (5π/3, 2π), f''(x) > 0 implies f(x) is concave upward.

Step 6: Check for asymptotes

Since the function is a product of two trigonometric functions without any division, there are no asymptotes. With all these features determined, you can sketch the function $$f(x) = \cos{x} \sin^2{x}$$ on the interval [0, 2π], taking into account its critical points, extrema, points of inflection, intervals of increasing and decreasing, and concavity.

Key Concepts

Graph Sketching

Graph sketching is all about visualizing the behavior of a function across its domain. When we sketch a graph, we aim to highlight key characteristics such as intercepts, extrema (like local maxima and minima), points of inflection, asymptotes, and intervals of increase or decrease. For the function \(f(x) = \cos{x} \sin^2{x}\), you start by identifying these key elements from calculus.
A well-sketched graph helps to convey much more than just the basic curve; it tells the story of a function’s rates of change and its general shape. Graph sketching combines mathematical analysis with geometric insight, creating an informative structure that reflects the function's true nature.
In this context, sketching \(f(x)\) involves finding its critical points, identifying where it increases or decreases, and the regions of concavity. The presence of trigonometric components also entails periodic behavior, which is essential to reflect accurately in the sketch within the given interval \[0, 2\pi\].

First and Second Derivatives

The essence of calculus is captured in derivatives, which measure change. The first derivative of a function, denoted \(f'(x)\), indicates the function's rate of change or slope at any point \(x\). When analyzing \(f(x) = \cos{x} \sin^2{x}\), the first derivative tells you where the function is increasing or decreasing.
The second derivative, \(f''(x)\), sheds light on the concavity of the function. It helps identify points of inflection where the function transitions from concave upward to concave downward, or vice versa. Calculating these derivatives generally involves applying product and chain rules, particularly for composite functions like those involving trigonometric terms.
For this specific \(f(x)\), we have:

• \(f'(x) = -\sin{x}\sin^2{x} + 2\cos^2{x}\sin{x}\)
• \(f''(x) = -2\sin{x}\cos{x}(3\cos^2{x} - 1)\)

These derivatives highlight not just the critical points but graph behavior at those points.

Critical Points

Critical points occur where the first derivative is zero or undefined, indicating potential maxima, minima, or points of inflection. Solving \(f'(x) = 0\) for the function \(f(x) = \cos{x} \sin^2{x}\) yields critical points on \[0, 2\pi\]: 0, \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), and \(2\pi\).
At each of these points, the function's behavior shifts drastically. To fluently determine whether these points are maxima or minima, one can use the first derivative test by checking signs before and after these points:

• If \(f'(x)\) transitions from positive to negative, the point is a local maximum.
• If it shifts from negative to positive, it's a local minimum.

In this exercise, \(x = \frac{\pi}{2}\) marks a local maximum, and \(x = \frac{3\pi}{2}\) a local minimum, which are vital in plotting an accurate graph.

Concavity and Inflection Points

Concavity refers to the direction the function curves, providing valuable insight into its shape. A function is concave upward if its second derivative is positive and concave downward if negative. This can change over intervals and is crucial for understanding how a function bends.
Points of inflection occur where concavity changes, corresponding to zeros of the second derivative \(f''(x)\). For \(f(x) = \cos{x} \sin^2{x}\), the equation \(-2\sin{x}\cos{x}(3\cos^2{x} - 1) = 0\) reveals points of inflection at \(x = 0\), \(\pi/3\), \(2\pi/3\), \(\pi\), \(4\pi/3\), \(5\pi/3\), and \(2\pi\). These guide local curve direction changes between regions of concavity:

• From \(0\) to \(\pi/3\), the function is concave upward.
• Between \(\pi/3\) and \(2\pi/3\), it is concave downward.

Such analysis enriches graph sketching with detail, clarifying fluid transitions within the targeted interval.