Problem 237 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

f is increasing on \([-\frac{3\pi}{4}, \frac{\pi}{4}]\), decreasing otherwise; local min at \(-\frac{3\pi}{4}\), max at \(\frac{\pi}{4}\); concave up on \([-\pi, -\frac{\pi}{2}]\) and \([\frac{\pi}{2}, \pi]\), concave down on \([-\frac{\pi}{2}, \frac{\pi}{2}]\); inflection points at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

Step 1: Find the First Derivative

To find the intervals where the function is increasing or decreasing, we need the first derivative of the function. Let \( f(x) = \sin(x)e^x \). Using the product rule, \( f'(x) = \sin(x)\frac{d}{dx}(e^x) + e^x\frac{d}{dx}(\sin(x)) = \sin(x) e^x + e^x \cos(x) \). Thus,\[ f'(x) = e^x(\sin(x) + \cos(x)) \].

Step 2: Determine Critical Points

Set the first derivative \( f'(x) = e^x(\sin(x) + \cos(x)) \) equal to zero to find critical points. Since \( e^x eq 0 \) for all \( x \), we solve \( \sin(x) + \cos(x) = 0 \). Let \( \frac{\sin(x)}{\cos(x)} = -1 \), providing \( \tan(x) = -1 \). The solutions to \( \tan(x) = -1 \) on \( x = [-\pi, \pi] \) are \( x = -\frac{3\pi}{4} \), \( x = \frac{\pi}{4} \).

Step 3: Test Intervals for Monotonicity

Evaluate \( f'(x) \) around the critical points to determine where \( f \) is increasing or decreasing. Choose test points such as \( x = -\pi, 0, \pi \). Compute:- \( f'(-\pi) = e^{-\pi}(\sin(-\pi) + \cos(-\pi)) = e^{-\pi}(-1) < 0 \) - \( f'(-\frac{3\pi}{4}) = 0 \) (critical point)- \( f'(0) = e^{0}(0+1) = 1 > 0 \)- \( f'(\frac{\pi}{4}) = 0 \) (critical point)- \( f'(\pi) = e^{\pi}(0 -1) < 0 \) Thus, \( f \) is decreasing on \( [-\pi, -\frac{3\pi}{4}] \) and \( [\frac{\pi}{4}, \pi] \). Increasing on \( [-\frac{3\pi}{4}, \frac{\pi}{4}] \).

Step 4: Find Local Minima and Maxima

Use the first derivative test at the critical points to evaluate local extrema:- At \( x = -\frac{3\pi}{4} \), \( f'(x) \) changes from negative to positive, indicating a local minimum.- At \( x = \frac{\pi}{4} \), \( f'(x) \) changes from positive to negative, indicating a local maximum.

Step 5: Find the Second Derivative

The second derivative helps determine concavity. Differentiate \( f'(x) \):\[ f''(x) = \frac{d}{dx}(e^x(\sin(x) + \cos(x))) = e^x(\sin(x) + \cos(x)) + e^x(\cos(x) - \sin(x)) \] Simplifying, \[ f''(x) = e^x(2\cos(x)) \].

Step 6: Determine Concavity

Evaluate \( f''(x) = 0 \) to find potential inflection points. Solve \( 2\cos(x) = 0 \), giving \( \cos(x) = 0 \). Solutions in \([-\pi, \pi]\) are \( x = -\frac{\pi}{2}, \frac{\pi}{2} \). - \( f(-\pi, -\frac{\pi}{2}) \) is concave up since \( f''(x) > 0 \).- \( f(-\frac{\pi}{2}, \frac{\pi}{2}) \) is concave down since \( f''(x) < 0 \).- \( f(\frac{\pi}{2}, \pi) \) is concave up since \( f''(x) > 0 \).

Step 7: Inflection Points

Inflection points occur where \( f''(x) \) changes sign. From the second derivative test, this occurs at \( x = -\frac{\pi}{2}, \frac{\pi}{2} \).

Key Concepts

First Derivative

The first derivative of a function is crucial for understanding how the function behaves. It lets us find intervals where a function is increasing or decreasing. To find the first derivative of the function \( f(x) = \sin(x) e^x \), we use the product rule. This rule applies when differentiating two functions that are multiplied together.

• Differentiate \( e^x \) to get \( e^x \).
• Differentiate \( \sin(x) \) to get \( \cos(x) \).
• Combine results: The first derivative \( f'(x) = \sin(x) e^x + \cos(x) e^x = e^x(\sin(x) + \cos(x)) \).

This derivative tells us how the slope of the function changes along its curve. By analyzing this slope, we can understand the behavior of the original function.

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. They are key since these points can indicate local maxima or minima. For \( f(x) = \sin(x)e^x \), we solve the equation:

• Set \( f'(x) = e^x(\sin(x) + \cos(x)) = 0 \).
• \( e^x eq 0 \) for all \( x \), so solve \( \sin(x) + \cos(x) = 0 \).
• This simplifies to \( \tan(x) = -1 \), providing solutions in the interval \([-\pi, \pi]\).
• Found critical points are \(-\frac{3\pi}{4} \) and \( \frac{\pi}{4} \).

These points give us potential locations of turning points on the graph. Testing around these helps confirm their nature.

Second Derivative

The second derivative of a function provides information about its concavity. Concavity describes how a function bends and curves. Calculating the second derivative involves differentiating the first derivative once more:

• Start with \( f'(x) = e^x(\sin(x) + \cos(x)) \).
• Differentiate to find \( f''(x) = e^x(2\cos(x)) \). Simplifying shows \( f''(x) = 2e^x\cos(x) \).

This derivative helps identify intervals of concavity and potential inflection points. A change in sign of \( f''(x) \) suggests an inflection point.

Concavity

Concavity tells us how a function is curved. If the second derivative \( f''(x) > 0 \), the function is concave up. If \( f''(x) < 0 \), the function is concave down. Analyzing \( f''(x) = 2e^x\cos(x) \):

• Solve \( f''(x) = 0 \) by finding when \( \cos(x) = 0 \).
• Solutions for \( x \) within \([-\pi, \pi]\) are \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).
• Evaluate the sign of \( f''(x) \) across the intervals divided by these points.

This reveals:

• Concave up on \([-\pi, -\frac{\pi}{2})\) and \((\frac{\pi}{2}, \pi]\).
• Concave down on \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Understanding concavity aids in sketching functions accurately.

Inflection Points

Inflection points are where a function changes concavity, representing a transition from concave up to concave down, or vice versa. They occur where the second derivative changes sign.
To find inflection points of \( f(x) = \sin(x)e^x \), observe when \( f''(x) = 2e^x\cos(x) \) changes sign:

• Identify zeroes of \( \cos(x) \) within \([-\pi, \pi]\), specifically at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• Verify sign change of \( f''(x) \) around these values to confirm inflection.

This means inflection points occur at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), where the function changes its bending.