Problem 236 - 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. \(\quad f(x)=\frac{\sin x}{x} \quad\) over \(\quad x=[-2 \pi, 2 \pi]\) \([2 \pi, 0) \cup(0,2 \pi]\)

Step-by-Step Solution

Answer

Use derivatives and tests to find intervals, extrema, and inflection points of \( f(x) = \frac{\sin x}{x} \). Assist with a graphing calculator.

Step 1: Find the first derivative

To determine the intervals where the function \( f(x) = \frac{\sin x}{x} \) is increasing or decreasing, we first find its first derivative \( f'(x) \). Using the quotient rule, we have:\[ f'(x) = \frac{x \cdot \cos x - \sin x}{x^2} \]

Step 2: Determine critical points

To find critical points where the function changes from increasing to decreasing (or vice versa), set the first derivative equal to zero:\[ x \cdot \cos x - \sin x = 0 \]Solve for \( x \) over the given domain \([-2\pi, 2\pi]\). This equation cannot be solved exactly by hand, so numerical methods or graphing may be needed to find approximate solutions.

Step 3: First derivative test

Apply the first derivative test around each critical point found to determine where the function is increasing or decreasing. Analyze each interval created by the critical points, using test points in each interval to check the sign of \( f'(x) \).

Step 4: Find the second derivative

To find intervals of concavity, compute the second derivative of \( f(x) \):Using the quotient rule again, we have:\[ f''(x) = \frac{(x^2(-\sin x) + 2x(\sin x - x \cos x))}{x^4} \] Simplify to make calculations easier for the next steps.

Step 5: Determine concavity

Find where \( f''(x) = 0 \) or undefined to determine where the function might change concavity. Use test points to assess the concavity over each interval divided by potential points of inflection.

Step 6: Second derivative test

Utilize the second derivative test on critical points identified in Step 2 to determine if these are local minima or maxima. Analyze the sign of \( f''(x) \) at these points. If \( f''(x) > 0 \), there's a local minimum; if \( f''(x) < 0 \), there's a local maximum.

Step 7: Identify inflection points

Evaluate the changes in concavity to identify inflection points, where the second derivative changes sign, indicating a shift from concave up to concave down or vice versa.

Step 8: Sketch the curve

Use the information gathered from the steps above to sketch the curve of \( f(x) = \frac{\sin x}{x} \). Highlight key features such as regions of increase/decrease, local maxima/minima, intervals of concavity, and inflection points.

Step 9: Calculator comparison

Check your hand calculations using a graphing calculator to visually confirm the intervals of increase, decrease, concavity, and location of local extrema and inflection points over \([-2\pi, 2\pi]\).

Key Concepts

First Derivative Test

The first derivative test is essential for understanding whether a function is increasing or decreasing. By calculating the derivative of a function, we determine where the function's slope is positive or negative. In the case of \( f(x) = \frac{\sin x}{x} \), the first step is to find its derivative:

• Using the quotient rule, we get \( f'(x) = \frac{x \cdot \cos x - \sin x}{x^2} \).

Set this derivative equal to zero to locate critical points, where the function transitions from increasing to decreasing or vice versa. We examine the sign of \( f'(x) \) in each interval around these critical points using test values:

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

The intervals of increase or decrease provide insight into the behavior of the function over the domain. If computational aids are required, especially for advanced functions, using a calculator for visual confirmation can be helpful.

Second Derivative Test

The second derivative test helps us determine whether a critical point is a local maximum, a local minimum, or possibly neither. After identifying critical points with the first derivative, we take the second derivative of the function:

• For \( f(x) = \frac{\sin x}{x} \), the second derivative is \[ f''(x) = \frac{(x^2(-\sin x) + 2x(\sin x - x \cos x))}{x^4} \].

Evaluate \( f''(x) \) at each critical point:

• If \( f''(x) > 0 \), there is a local minimum at that point.
• If \( f''(x) < 0 \), there is a local maximum at that point.
• If \( f''(x) = 0 \), the test is inconclusive, and other methods may be necessary.

The sign of the second derivative tells you about the "curvature" of the function: whether it curves upwards or downwards near the critical points.

Critical Points

Critical points of a function occur where the first derivative is zero or undefined, indicating potential local maxima, minima, or points of horizontal tangency. For the function \( f(x) = \frac{\sin x}{x} \), we determined the derivative and solved:

• \( f'(x) = \frac{x \cdot \cos x - \sin x}{x^2} = 0 \)

Since solving this analytically can be challenging, numerical approximation or plotting might be needed. These points are crucial as they divide the domain into intervals, highlighting areas to apply the first derivative test:

• Assessing \( f'(x) \) around these points helps identify increasing or decreasing intervals.

Critical points are foundational in sketching the function's graph because they often mark the "turning points," where the graph changes direction.

Concavity

Concavity describes how the graph of a function "bends." It informs us whether a function opens upwards or downwards. The second derivative, \( f''(x) \), determines concavity:

• If \( f''(x) > 0 \), the function is concave up (like a cup).
• If \( f''(x) < 0 \), the function is concave down (like a frown).

For \( f(x) = \frac{\sin x}{x} \), the concavity depends on the sign of\[ f''(x) = \frac{(x^2(-\sin x) + 2x(\sin x - x \cos x))}{x^4} \].Make use of test points in each interval set out by finding where \( f''(x) = 0 \) or undefined:

• This gives insight into the interval-wide behavior of the function's graph.

Understanding concavity is crucial for sketching a more precise and refined graph.

Inflection Points

Inflection points are where a function changes concavity, switching from concave up to concave down or vice versa. These points occur where the second derivative changes sign.

• The core idea is that at an inflection point, \( f''(x) \) transitions from positive to negative or from negative to positive.

In the context of \( f(x) = \frac{\sin x}{x} \), solving for where the simplified form of \( f''(x) \) equals zero or does not exist is crucial:

• These locations suggest potential inflection points.
• Verify by checking the sign change on either side.

Inflection points provide detailed shape changes in the graph, informing whether sections of the graph are curving differently. These points give additional accuracy in sketching graphs, adding more visual clarity.