Problem 22 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=\tan (x)-x,-\pi

Step-by-Step Solution

Answer

The function is concave up on \((0, \pi/2) \cup (-\pi, -\pi/2)\) and concave down on \((-\pi/2, 0) \cup (\pi/2, \pi)\). No points of inflection or local extrema exist.

Step 1: Find the First Derivative

To start, differentiate the function. Given \(f(x) = \tan(x) - x\), the first derivative is \(f'(x) = \sec^2(x) - 1\).

Step 2: Identify Critical Points

Critical points occur where \(f'(x) = 0\) or where \(f'(x)\) is undefined. Set \(\sec^2(x) - 1 = 0\). Solving for \(x\), we get \[ \sec^2(x) = 1 \] \[ \cos^2(x) = 1 \]Thus, \(x = n\pi\), where \(n\) is an integer. Within the interval \(-\pi < x < \pi\), \(x = 0\) is the only critical point.

Step 3: Find the Second Derivative

Differentiate the first derivative \(f'(x) = \sec^2(x) - 1\) to get the second derivative. The second derivative is \[ f''(x) = 2 \sec^2(x)\tan(x) \].

Step 4: Determine Concavity

The function is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\).* \(f''(x) > 0\) implies \(\tan(x) > 0\), i.e., in the intervals \(0 < x < \pi/2\) and \(-\pi < x < -\pi/2\). * \(f''(x) < 0\) implies \(\tan(x) < 0\), i.e., in the intervals \(-\pi/2 < x < 0\) and \( \pi/2 < x < \pi\).

Step 5: Identify Points of Inflection

Points of inflection occur where the concavity changes sign, corresponding to solutions of \(f''(x) = 0\). The expression \(f''(x)\) never equals zero since \(2 \sec^2(x)\tan(x) eq 0\) for any \(x\). Thus, there are no points of inflection on the interval \(-\pi < x < \pi\).

Step 6: Classify Critical Points

Use the second derivative test to classify critical points. Since \(f''(0) = 0\), the test is inconclusive at \(x = 0\) for identifying maxima or minima.By analyzing the behavior of the function, \(x = 0\) is neither a local maximum nor a local minimum.

Key Concepts

First Derivative

The first derivative of a function provides essential information about its slope or rate of change. In the case of the function \( f(x) = \tan(x) - x \), we find the first derivative by differentiating:

• \( f'(x) = \sec^2(x) - 1 \)

This derivative informs us about the intervals where the function is increasing or decreasing and helps identify the critical points. For a function:

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

The first derivative sets the stage for many future analyses, including finding the critical points.

Second Derivative

The second derivative of a function tells us about the concavity, which is the curvature of the function's graph. Differentiating the first derivative, \( f'(x) = \sec^2(x) - 1 \), gives us:

• \( f''(x) = 2 \sec^2(x)\tan(x) \)

• If \( f''(x) > 0 \), the function is concave up, resembling a "cup" shape.
• If \( f''(x) < 0 \), the function is concave down, resembling a "cap" shape.

This information is crucial for identifying regions of different slopes and behaviors across the graph. It helps to understand how steep or flat a function is and whether intervals exhibit a concave up or down shape.

Critical Points

Critical points occur where the first derivative is zero or undefined. For the function \( f(x) = \tan(x) - x \):

• Solve \( \sec^2(x) - 1 = 0 \) leading to \( \cos^2(x) = 1 \), so \( x = n\pi \), where \( n \) is an integer.

Within the interval \( -\pi < x < \pi \), the only critical point is \( x = 0 \).

• Critical points are locations where the graph may have a peak, valley, or a point of horizontal tangent.
• They play a vital role in analyzing the function's local behavior.

To fully understand whether these points are maxima or minima, further testing with the second derivative is usually required.

Points of Inflection

Points of inflection are where a function changes from being concave up to concave down, or vice versa. To locate them, we look for values where the second derivative equals zero and changes signs. For our function:

• The second derivative \( f''(x) = 2 \sec^2(x)\tan(x) \) never equals zero in the interval \(-\pi < x < \pi\).

Thus, there are no points of inflection for this particular function in the given interval.

• Points of inflection are those interesting spots where the graph's curvature reverses, offering insight into its changing shape.
• They are especially useful for understanding the overall behavior of the function as it evolves.

In this case, the absence of such points simplifies the concavity analysis.