Problem 10 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each of the following functions, determine where the function is increasing and where it is decreasing. Find the \(x\) -coordinates of all local maxima and minima. (Give exact answers, not numerical approximations.) (a) \(f(x)=2 x^{3}-24 x+4\) (b) \(f(x)=x^{3}-3 x^{2}-9 x+2\)

Step-by-Step Solution

Answer

The function (a) increases on the interval \(-2

Step 1: Find the first derivative of the functions

Differentiate the function to find the critical points. For (a) \(f'(x)= 6x^{2}-24\), for (b) \(f'(x)= 3x^{2}-6x-9\)

Step 2: Find the critical points

The critical points are the solutions of the derivative equal to zero. For (a) Solve \(6x^{2}-24=0\) to get \(x=-2,2\). For (b) Solve \(3x^{2}-6x-9 = 0\) using the quadratic formula to obtain \(x = -1, 3\)

Step 3: Determine increasing or decreasing intervals, maxima and minima

Choose a test point in the intervals determined by the critical points and substitute into the first derivative. If the result is positive the function is increasing, if negative it is decreasing. A change from increasing to decreasing indicates a local maximum, and a change from decreasing to increasing indicates a local minimum. For (a) the function is decreasing in the interval \(-\infty, -2]\), increasing in the interval \(-2, 2\), and decreasing again in the interval \([2, \infty)\). Therefore, there is a local minimum at \(x=-2\) and a local maximum at \(x=2\). For (b) the function is increasing in the interval \(-\infty, -1]\), decreasing in the interval \(-1, 3\), and increasing again in the interval \([3, \infty)\). So, there is a local maximum at \(x=-1\) and a local minimum at \(x=3\)

Step 4: Summarization

The function (a) has a local minimum at \(x=-2\) and a local maximum at \(x=2\). The function (b) has a local maximum at \(x=-1\) and a local minimum at \(x=3\)

Key Concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), provides crucial insight into a function's behavior. By differentiating a given function, we essentially determine the rate at which the function's value is changing at any point \( x \). This is key for analyzing trends—whether the function is increasing or decreasing—and is foundational for finding critical points.

When we differentiate a polynomial like \( f(x) = 2x^3 - 24x + 4 \), we apply the power rule: bringing down the exponent and reducing it by one. For this function, the derivative is \( f'(x) = 6x^2 - 24 \). Similarly, for \( f(x) = x^3 - 3x^2 - 9x + 2 \), the derivative is \( f'(x) = 3x^2 - 6x - 9 \). With the first derivative found, we are a step closer to understanding the function's graphical characteristics.

Critical Points

Critical points of a function occur where its first derivative equals zero or is undefined. These points indicate potential locations for extreme values—maxima or minima. To find the critical points, we solve \( f'(x) = 0 \). For the function \( f(x) = 2x^3 - 24x + 4 \), solving \( 6x^2 - 24 = 0 \) gives critical points at \( x = -2 \) and \( x = 2 \).

For \( f(x) = x^3 - 3x^2 - 9x + 2 \), we apply the quadratic formula on \( 3x^2 - 6x - 9 = 0 \) to find \( x = -1 \) and \( x = 3 \). Critical points are pivotal in determining where the function might transition from increasing to decreasing or vice versa, revealing the function's peaks and valleys.

Increasing and Decreasing Intervals

Once the critical points are known, we use them to break the number line into intervals. The behavior of a function can differ in each interval. By testing points within these intervals using the derivative \( f'(x) \), we determine the function's inclination.

For example, consider the function \( f(x) = 2x^3 - 24x + 4 \). Test a point from each interval:

• In interval \((-\infty, -2)\), if \( f'(x) \) is negative, the function is decreasing.
• In interval \((-2, 2)\), if \( f'(x) \) is positive, the function is increasing.
• In interval \((2, \infty)\), if \( f'(x) \) is negative, the function is decreasing.

Similarly, analyze \( f(x) = x^3 - 3x^2 - 9x + 2 \) with points from each interval determined by \( x = -1 \) and \( x = 3 \). Understanding these intervals helps graph the function accurately, revealing where it rises or falls along the x-axis.

Local Maxima and Minima

Local maxima and minima are specific types of critical points where the function attains peak values (highs) or troughs (lows), but not necessarily the absolute maximum or minimum.

To identify these points, observe where the function's derivative changes sign. For \( f(x) = 2x^3 - 24x + 4 \), a sign change from positive to negative at \( x = 2 \) indicates a local maximum, whereas a change from negative to positive at \( x = -2 \) shows a local minimum.

For \( f(x) = x^3 - 3x^2 - 9x + 2 \):

• A negative to positive change at \( x = 3 \) signifies a local minimum.
• A positive to negative change at \( x = -1 \) implies a local maximum.

These insights enable better understanding and representation of the function on a graph, pinpointing key points where the function's slope drastically shifts, marking significant changes in behavior.

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Problem 10

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Problem 11

Other exercises in this chapter

Problem 10

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute

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Problem 10

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critic

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Problem 11

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute

View solution

Problem 11

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critic

View solution

Key Concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), provides crucial insight into a function's behavior. By differentiating a given function, we essentially determine the rate at which the function's value is changing at any point \( x \). This is key for analyzing trends—whether the function is increasing or decreasing—and is foundational for finding critical points.

When we differentiate a polynomial like \( f(x) = 2x^3 - 24x + 4 \), we apply the power rule: bringing down the exponent and reducing it by one. For this function, the derivative is \( f'(x) = 6x^2 - 24 \). Similarly, for \( f(x) = x^3 - 3x^2 - 9x + 2 \), the derivative is \( f'(x) = 3x^2 - 6x - 9 \). With the first derivative found, we are a step closer to understanding the function's graphical characteristics.

Critical Points

Critical points of a function occur where its first derivative equals zero or is undefined. These points indicate potential locations for extreme values—maxima or minima. To find the critical points, we solve \( f'(x) = 0 \). For the function \( f(x) = 2x^3 - 24x + 4 \), solving \( 6x^2 - 24 = 0 \) gives critical points at \( x = -2 \) and \( x = 2 \).

For \( f(x) = x^3 - 3x^2 - 9x + 2 \), we apply the quadratic formula on \( 3x^2 - 6x - 9 = 0 \) to find \( x = -1 \) and \( x = 3 \). Critical points are pivotal in determining where the function might transition from increasing to decreasing or vice versa, revealing the function's peaks and valleys.

Increasing and Decreasing Intervals

Once the critical points are known, we use them to break the number line into intervals. The behavior of a function can differ in each interval. By testing points within these intervals using the derivative \( f'(x) \), we determine the function's inclination.

For example, consider the function \( f(x) = 2x^3 - 24x + 4 \). Test a point from each interval:

• In interval \((-\infty, -2)\), if \( f'(x) \) is negative, the function is decreasing.
• In interval \((-2, 2)\), if \( f'(x) \) is positive, the function is increasing.
• In interval \((2, \infty)\), if \( f'(x) \) is negative, the function is decreasing.

Similarly, analyze \( f(x) = x^3 - 3x^2 - 9x + 2 \) with points from each interval determined by \( x = -1 \) and \( x = 3 \). Understanding these intervals helps graph the function accurately, revealing where it rises or falls along the x-axis.

Local Maxima and Minima

Local maxima and minima are specific types of critical points where the function attains peak values (highs) or troughs (lows), but not necessarily the absolute maximum or minimum.

To identify these points, observe where the function's derivative changes sign. For \( f(x) = 2x^3 - 24x + 4 \), a sign change from positive to negative at \( x = 2 \) indicates a local maximum, whereas a change from negative to positive at \( x = -2 \) shows a local minimum.

For \( f(x) = x^3 - 3x^2 - 9x + 2 \):

• A negative to positive change at \( x = 3 \) signifies a local minimum.
• A positive to negative change at \( x = -1 \) implies a local maximum.

These insights enable better understanding and representation of the function on a graph, pinpointing key points where the function's slope drastically shifts, marking significant changes in behavior.