Problem 17 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The graph of \(f^{\prime}\left(\right.\) not \(f\), but \(f^{\prime}\) ) is a parabola with \(x\) -intercepts of \(-\pi\) and \(2 \pi\) and a \(y\) -intercept of \(-2\). (a) Draw a graph of \(f^{\prime}\). (b) Write an equation for \(f^{\prime}\). This equation should have no unknown constants. (c) On the graph you drew in part (a), go back and label the \(x\) - and \(y\) -coordinates of the vertex. (d) Find \(f^{\prime \prime}(x)\). (e) This part of the question asks about \(f\), not \(f^{\prime}\). i. Where does \(f\) have a local maximum? Explain your reasoning clearly and briefly. ii. Where does \(f\) have a local minimum? Explain your reasoning clearly and briefly. iii. Does \(f\) have an absolute maximum or minimum value? Explain. iv. The function \(f\) has a single point of inflection. What is the \(x\) -coordinate of this point of inflection? Suppose you are told that the \(y\) -coordinate of the point of inflection is \(-1\). Find the equation of the tangent line to the graph of \(f\) at its point of inflection.

Step-by-Step Solution

Answer

The derivative function \(f^{\prime}(x) = -1/\pi^2 (x-\pi/2)^2-2\), its second derivative \(f^{\prime \prime}(x) = -2/\pi^2 (x-\pi/2)\). The function \(f\) has a local maximum at \(-\pi\), a local minimum at \(2\pi\), no absolute extrema, and a point of inflection at \(\pi/2\) with the tangent line \(y = -1\).

Step 1: Draw the Graph of the Derivative

Given the \(x\)-intercepts are \(-\pi\) and \(2 \pi\), and the \(y\)-intercept is \(-2\), we can sketch a downward-opening parabola with these intercepts.

Step 2: Write an Equation for the Derivative

The standard form for the equation of a parabola is \(y=a(x-h)^2+k\), where \((h, k)\) are the coordinates of the vertex. We know that the parabola opens downwards, so \(a\) is negative. The x-coordinate of the vertex is the average of the two \(x\)-intercepts: \(h = (\(-\pi + 2\pi)/2 = \pi/2\). The equation of the derivative is \(f^{\prime}(x) = a(x-\pi/2)^2-2\). Since it passes through the point \((2 \pi, 0)\), we get \(a = -1/\pi^2\), so \(f^{\prime}(x) = -1/\pi^2 (x-\pi/2)^2-2\).

Step 3: Mark the Vertex on the Graph

The vertex of the parabola occurs at \((\pi/2, -2)\). This point should be labelled on the graph.

Step 4: Find the Second Derivative

Differentiating \(-1/\pi^2 (x-\pi/2)^2-2\) with respect to \(x\), \(f^{\prime \prime}(x) = -2/\pi^2 (x-\pi/2)\).

Step 5: Analyze the Function Based on its Derivatives

i. A local maximum of \(f\) must occur where \(f^{\prime}\) changes from positive to negative. Here, looking at the graph of \(f^{\prime}\), this can be seen at \(x=-\pi\). \ii. A local minimum of \(f\) occurs where \(f^{\prime}\) changes from negative to positive. This is at \(x=2\pi\). \iii. \(f'\) has no local extremum. Therefore, \(f\) has neither absolute maximum nor minimum. \iv. The point of inflection \(x\)-coordinate is at the root of \(f^{\prime\prime}(x) = 0\). So \(x=\pi/2\). The tangent line at this point can be found using the point-slope formula \(y-y_1 = m(x-x_1)\) where \(m = f^{\prime}(x_1)\) and \((x_1, y_1)\) is the point of inflection, yielding the equation \(y + 1 = 0(x-\pi/2)\). Simplifying this gives \(y = -1\).

Key Concepts

Parabola

A parabola is a U-shaped curve that can open upwards or downwards. It is the graph of a quadratic function, typically expressed in the standard form \( y = ax^2 + bx + c \), or the vertex form \( y = a(x-h)^2 + k \). The vertex form is useful when you want to identify the vertex of the parabola easily, where \( (h,k) \) is the vertex's location.
The direction in which the parabola opens depends on the sign of the coefficient \( a \):

• If \( a \) is positive, the parabola opens upwards.
• If \( a \) is negative, the parabola opens downwards.

In the given exercise, we have a downward-opening parabola for the derivative function \( f^{\prime}(x) \), confirmed by its negative \( a \)-value. Essential attributes of the parabola include \( x \)-intercepts and the \( y \)-intercept, which help in sketching the graph and finding the equation of the parabola.

Critical Points

Critical points occur in a function where its derivative is zero or undefined. These are significant locations, as they help identify where a function might have maximum or minimum values or changes in behavior.
For a function \( f(x) \), you can find the critical points by solving \( f^{\prime}(x) = 0 \). In this exercise, it is especially relevant for identifying where \( f \) might have local maxima or minima, by examining \( f^{\prime}(x) \).

• If \( f^{\prime}(x) = 0 \) and the sign of \( f^{\prime} \) changes from positive to negative, you have a local maximum.
• If \( f^{\prime}(x) = 0 \) and the sign changes from negative to positive, there is a local minimum.

Critical points do not guarantee local extrema; instead, they indicate where to look using the graph's behavior. In this context, the critical points provide insight into \( f \) possibly achieving local extremum values at specific \( x \) values.

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function. For a function \( f(x) \), its derivative \( f^{\prime}(x) \) reflects how \( f(x) \) changes with respect to \( x \). This first derivative is pivotal in determining key features of a function like slope and the location of critical points.
In the context of the exercise, the derivative \( f^{\prime}(x) \) is given as a parabola. By analyzing this derivative, we can infer properties of the original function \( f(x) \), such as its maximum, minimum, or inflection points.

• A positive \( f^{\prime}(x) \) means \( f(x) \) is increasing in that interval.
• A negative \( f^{\prime}(x) \) suggests \( f(x) \) is decreasing.

The second derivative \( f^{\prime\prime}(x) \) offers further insights, helping identify concavity and possible inflection points. Differentiating the first derivative gives \( f^{\prime\prime}(x) \), indicating changes in the slope of \( f(x) \).

Inflection Point

An inflection point on a curve is where the concavity changes - from concave up to concave down, or vice versa. It's where the second derivative \( f^{\prime\prime}(x) \) changes sign or equals zero. Identifying these points helps to understand the overall shape and behavior of the graph.
In this exercise, we find the \( x \)-coordinate of the inflection point by solving \( f^{\prime\prime}(x) = 0 \). Once we have this coordinate, corroborating with the known \( y \)-coordinate helps in determining the tangent line at this inflection point.

• For \( f \) to have an inflection point at \( x = \rac{\pi}{2} \), the concavity must change around this value.
• The equation of the tangent line at this point provides more precise information regarding the behavior of \( f \) around this shift.

Understanding inflection points is key to graph sketching and predicting how the curve transitions between different portions.