Problem 22 - Applied Calculus

Question

For the functions in problems, do the following: (a) Find \(f^{\prime}\) and \(f^{\prime \prime}\). (b) Find the critical points of \(f\). (c) Find any inflection points of \(f\). (d) Evaluate \(f\) at its critical points and at the endpoints of the given interval. Identify local and global maxima and minima of \(f\) in the interval. (e) Graph \(f\). $$ f(x)=e^{-x} \sin x \quad(0 \leq x \leq 2 \pi) $$

Step-by-Step Solution

Answer

Critical points are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). Local max at \( x = \frac{\pi}{4} \), min at \( x = \frac{5\pi}{4} \). Inflection points are the same.

Step 1: Differentiate to find First Derivative

To find the first derivative \( f'(x) \), use the product rule. For \( f(x) = e^{-x} \sin x \), the derivative is:\[ f'(x) = (e^{-x})' \sin x + e^{-x} (\sin x)' \]Calculate each part separately:\[ (e^{-x})' = -e^{-x} \quad \text{and} \quad (\sin x)' = \cos x \]So,\[ f'(x) = -e^{-x} \sin x + e^{-x} \cos x = e^{-x} (\cos x - \sin x) \]

Step 2: Differentiate to find Second Derivative

Now, find the second derivative \( f''(x) \) using the product rule again:\[ f''(x) = (e^{-x} ( \cos x - \sin x ))' \]Apply the derivative:\[ f''(x) = (e^{-x})' (\cos x - \sin x) + e^{-x} (\cos x - \sin x)' \]Which results in:\[ f''(x) = -e^{-x} (\cos x - \sin x) + e^{-x} (-\sin x - \cos x) = e^{-x} (-2\cos x + 2\sin x) \]

Step 3: Find Critical Points

Critical points occur where \( f'(x) = 0 \). Set \( e^{-x}(\cos x - \sin x) = 0 \). Since \( e^{-x} eq 0 \), the expression reduces to:\[ \cos x - \sin x = 0 \rightarrow \cos x = \sin x \]This implies \( x = \frac{\pi}{4} + k\pi \). In the interval \( [0, 2\pi] \), the critical points are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Step 4: Find Inflection Points

Inflection points occur where \( f''(x) = 0 \). Set \( e^{-x} (-2\cos x + 2\sin x) = 0 \). Again, as \( e^{-x} eq 0 \), we simplify to:\[ -2\cos x + 2\sin x = 0 \rightarrow \cos x = \sin x \]Again, solving within the interval \( [0, 2\pi] \), we get the same points: \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). These are also inflection points.

Step 5: Evaluate at Critical Points and Endpoints

Evaluate \( f(x) \) at \( x = 0, \frac{\pi}{4}, \frac{5\pi}{4}, 2\pi \):- \( f(0) = e^{0} \sin(0) = 0 \)- \( f\left(\frac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2e^{\frac{\pi}{4}}} \)- \( f\left(\frac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2e^{\frac{5\pi}{4}}} \)- \( f(2\pi) = e^{-2\pi} \sin(2\pi) = 0 \)\( f\left(\frac{\pi}{4}\right) \) is a local maximum, and \( f\left(\frac{5\pi}{4}\right) \) is a local minimum. Global extrema should be checked numerically or graphically.

Step 6: Graph the Function

Using graphing software or hand-plotting, sketch \( f(x) = e^{-x} \sin x \) over the interval \( [0, 2\pi] \). It should show peaks at \( x = \frac{\pi}{4} \) and valleys at \( x = \frac{5\pi}{4} \), with the function approaching zero at \( x = 0 \) and \( x = 2\pi \).

Key Concepts

Derivative

In calculus, the derivative of a function is a measure of how the function changes as its input changes. Derivatives are fundamental to understanding the behavior of functions. When you differentiate a function, you apply rules that calculate its rate of change. The derivative, denoted as \(f'(x)\), tells us how steep the graph of the function is at any particular point.

• The first derivative \(f'(x)\) helps identify increasing and decreasing intervals of the function.
• It's also used to find critical points by setting \(f'(x) = 0\).

In the given problem, we used the product rule, which applies when differentiating products of functions. If \(f(x) = e^{-x} \sin x\), the product rule tells us:

• Differentiating \(e^{-x}\) gives \(-e^{-x}\).
• Differentiating \(\sin x\) gives \(\cos x\).
• Combining these, we get \(f'(x) = e^{-x}(\cos x - \sin x)\).

Understanding the derivatives allows us to explore more about the function's shape and behavior.

Critical Points

Critical points of a function are where the derivative \(f'(x)\) is zero or undefined. These points are crucial as they often correspond to local maxima or minima, where the function changes its direction.

• Local maxima are points where the function values are higher than all nearby points.
• Local minima are points where the function values are lower than all nearby points.

To find critical points in our example, we solve \(e^{-x}(\cos x - \sin x) = 0\). Since \(e^{-x} eq 0\), the equation simplifies to \(\cos x = \sin x\), leading to solutions at \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\). In these locations, the function has local extrema.After locating critical points, we often classify them by examining the second derivative or using tests like the first derivative test for increasing or decreasing behavior at these points.

Inflection Points

Inflection points are points on a curve where the concavity changes. This means the curve shifts from concave up (like a cup) to concave down (like a frown), or vice versa. To find inflection points, we set \(f''(x) = 0\), solving for points where the second derivative changes sign.

• Concave up regions are where the graph is shaped like a cup \((\cap)\).
• Concave down regions resemble the shape of a frown \((\cup)\).

In our problem, we found \(f''(x) = e^{-x}(-2\cos x + 2\sin x)\). Setting this to zero leads us back to the same critical points: \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\). At these points, we checked whether the concavity indeed changes, confirming these are inflection points.

Function Graphing

Graphing a function provides a visual understanding of its behavior across a defined interval. When plotting the function \(f(x) = e^{-x} \sin x\), several features help guide how it looks:

• Begin by plotting points at critical and inflection points, like \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).
• These will manifest as peaks and valleys in the graph.
• Consider endpoints of the interval, \(x = 0\) and \(x = 2\pi\), where the function returns to zero.

Using software or manual plotting, the function's graph should show the expected peaks at the critical points. The values of the function at endpoints and critical points, as calculated, guide the sketch of the curve. This helps not only with visual analysis but also a practical check on where local maxima, minima, and inflections might occur.