Problem 237 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f} \\ {\text { c. intervals where } f \text { is concave up and concave }} \\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . Sketch the curve, then }} \\ {\text { use a calculator to compare your answer. If you }} \\\ {\text { cannot determine the exact answer analytically, use }} \\ {\text { a calculator. }}\end{array} $$ $$ f(x)=\sin (x) e^{x} \text { over } x=[-\pi, \pi] $$

Step-by-Step Solution

Answer

The function is increasing on \([-\frac{3\pi}{4}, \frac{\pi}{4}]\) with a local minimum at \(x=-\frac{3\pi}{4}\) and a local maximum at \(x=\frac{\pi}{4}\). Concave up on \([-\frac{\pi}{2}, \frac{\pi}{2}]\) with inflection points at \(x=\pm\frac{\pi}{2}\).

Step 1: Find the First Derivative

First we need to find the derivative of the function to determine where it is increasing or decreasing. Use the product rule to differentiate the function. Given the function \( f(x) = \sin(x) e^x \), we differentiate as follows:\( f'(x) = \cos(x)e^x + \sin(x)e^x = e^x(\cos(x) + \sin(x)) \).

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points:\[ e^x(\sin(x) + \cos(x)) = 0 \]Since \( e^x eq 0 \) for any real \( x \), solve \( \sin(x) + \cos(x) = 0 \). This simplifies to \( \tan(x) = -1 \), giving solutions \( x = -\frac{3\pi}{4}, \frac{\pi}{4} \) within the interval \([-\pi, \pi]\).

Step 3: Determine Intervals of Increase/Decrease

Examine the sign of \( f'(x) = e^x(\sin(x) + \cos(x)) \) in the intervals created by critical points:- On \([-\pi, -\frac{3\pi}{4}]\) and \([\frac{\pi}{4}, \pi]\), \( \sin(x) + \cos(x) < 0 \). Hence, \( f(x) \) is decreasing.- On \([-\frac{3\pi}{4}, \frac{\pi}{4}]\), \( \sin(x) + \cos(x) > 0 \). Hence, \( f(x) \) is increasing.

Step 4: Identify Local Minima/Maxima

Use the first derivative test:- At \( x = -\frac{3\pi}{4} \), \( f(x) \) changes from decreasing to increasing, indicating a local minimum.- At \( x = \frac{\pi}{4} \), \( f(x) \) changes from increasing to decreasing, indicating a local maximum.

Step 5: Find the Second Derivative

To analyze concavity, compute the second derivative of \( f(x) \). Differentiate \( f'(x) = e^x(\sin(x) + \cos(x)) \):\[ f''(x) = e^x(\sin(x) + \cos(x)) + e^x(\cos(x) - \sin(x)) = 2e^x\cos(x) \].

Step 6: Determine Concavity Intervals

Find where \( f''(x) = 2e^x\cos(x) \) changes sign:- \( \cos(x) = 0 \) when \( x = -\frac{\pi}{2}, \frac{\pi}{2} \) within \([-\pi, \pi]\). Evaluate the sign between intervals:- On \([-\pi, -\frac{\pi}{2}]\), \( \cos(x) < 0 \) so \( f(x) \) is concave down.- On \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \( \cos(x) > 0 \) so \( f(x) \) is concave up.- On \([\frac{\pi}{2}, \pi] \), \( \cos(x) < 0 \) so \( f(x) \) is concave down.

Step 7: Identify Inflection Points

Inflection points occur where \( f''(x) \) changes sign:- At \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), \( f''(x) \) changes sign, indicating inflection points at these values.

Step 8: Sketch and Verify

Sketch the graph of \( f(x) \) based on found intervals and points. Use a calculator to ensure results. Confirm increasing/decreasing behavior, concavity, and local extrema against your sketch and calculator output.

Key Concepts

Derivative

In calculus, the derivative of a function provides us with crucial insights about the function's rate of change. For the function \( f(x) = \sin(x) e^x \), finding the derivative is essential for understanding its increasing or decreasing nature. When you differentiate, you use the product rule in this case, as the function is a product of \( \sin(x) \) and \( e^x \).The derivative is calculated as: \[ f'(x) = \cos(x)e^x + \sin(x)e^x = e^x(\cos(x) + \sin(x))\]This expression shows how the function's values change across different \( x \) values. Since \( e^x \) is always positive, the sign of \( \cos(x) + \sin(x) \) determines whether the function is increasing or decreasing.

Critical Points

Critical points are crucial for understanding the behavior of functions. They occur where the derivative \( f'(x) \) is zero or undefined. In our example, by setting the derivative \( e^x(\sin(x) + \cos(x)) = 0 \), we focus on when \( \sin(x) + \cos(x) = 0 \), since \( e^x eq 0 \) for any real number \( x \).For the interval \([ -\pi, \pi ]\), this simplifies to solving \( \tan(x) = -1 \), which gives critical points at \( x = -\frac{3\pi}{4} \) and \( x = \frac{\pi}{4} \). These points are where potential changes in the direction of the function's graph might occur, indicating possible local minima or maxima.

Increasing Decreasing Intervals

Determining where a function is increasing or decreasing involves examining the sign of its derivative, \( f'(x) \). In our problem, we analyzed intervals based on critical points:

• On \([ -\pi, -\frac{3\pi}{4} ]\) and \([ \frac{\pi}{4}, \pi ]\), since \( \sin(x) + \cos(x) < 0 \), the function is decreasing.
• On \([ -\frac{3\pi}{4}, \frac{\pi}{4} ]\), \( \sin(x) + \cos(x) > 0 \), so the function is increasing.

These intervals tell us where the function is climbing upwards or falling downwards on the graph, providing a clear picture of the function's behavior over the specified range.

Concavity

The concept of concavity explains the bending nature of the graph. A function is concave up if its graph opens upwards, like a cup or smile, and concave down if it opens downwards. To explore concavity, we need the second derivative, \( f''(x) \).For our function, we compute:\[f''(x) = e^x(\sin(x) + \cos(x)) + e^x(\cos(x) - \sin(x)) = 2e^x \cos(x) \]Examining \( 2e^x \cos(x) \), the sign of \( \cos(x) \) determines concavity. Within \([ -\pi, \pi ]\):

• On \([ -\pi, -\frac{\pi}{2} ]\) and \([ \frac{\pi}{2}, \pi ]\), since \( \cos(x) < 0 \), the function is concave down.
• On \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\), \( \cos(x) > 0 \), so the function is concave up.

Understanding concavity helps us see where the graph changes shape and assists in verifying inflection points.

Inflection Points

Inflection points occur where there is a change in the concavity of the function. It’s where the graph switches from being concave up to concave down or vice versa. They are found where the second derivative, \( f''(x) \), changes sign.In our function, \( f''(x) = 2e^x \cos(x) \), changes sign at:

• \( x = -\frac{\pi}{2} \)
• \( x = \frac{\pi}{2} \)

These are the inflection points for \( f(x) \) as the concavity transitions at these \( x \) values. Identifying inflection points is vital for graph sketching, as they mark the spots where the nature of the curve changes.