Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

The first derivative \(f^{\prime}\) is given. Find all values of \(x\) that make the function \(f(a)\) a local minimum and \((b)\) a local maximum. \(f^{\prime}(x)=(x-1)^{2}(x-2)^{2}(x-3)^{2}(x-4)^{2}\)

Step-by-Step Solution

Answer

There are no local minima or maxima points for the given function.

Step 1: Identify Critical Points

To find critical points, set the derivative \( f'(x) \) equal to zero: \((x-1)^2 (x-2)^2 (x-3)^2 (x-4)^2 = 0\). The critical points are where each factor equals zero, which gives us \( x = 1, 2, 3, 4 \).

Step 2: Analyze the Nature of Critical Points

Since \((x-1)^2, (x-2)^2, (x-3)^2,\) and \((x-4)^2\) are all even powers, the function \(f'(x)\) changes concavity at each of these points without switching signs – hence, there is no sign change through these critical points. The value of \(f'(x)\) remains non-negative for all \(x\), therefore no critical point corresponds to a local extremum under these circumstances.

Step 3: Conclude the Analysis

Because the derivative \(f'(x)\) does not change sign across any of the critical points, there is neither local minima nor local maxima at these points. The function just flattens without any change from increasing to decreasing or vice versa at these critical points.

Key Concepts

First Derivative

The first derivative of a function, denoted as \( f'(x) \), is a tool that tells us the rate of change or the slope of the function \( f(x) \) at any point \( x \). It is found by differentiating the function regarding \( x \). In the given problem, the first derivative \( f'(x) \) is written succinctly using factored form:

• \( f'(x) = (x-1)^2 (x-2)^2 (x-3)^2 (x-4)^2 \)

This factored structure helps in identifying the critical points easily when the derivative is set to zero. Critical points arise because each factor can be zero, revealing potential spots where local extrema could exist. Understanding the behavior of these factors at a given \( x \) is crucial, especially when they have even powers, as it can affect the nature of any extremum present.

Local Extrema

Local extrema refer to local minimum or maximum values of a function. These are points where a function changes its direction: from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). To find local extrema, you typically:

• Compute and solve the first derivative \( f'(x) = 0 \) to identify critical points.
• Use the first derivative test around those points.

In the problem at hand, we have critical points at \( x = 1, 2, 3, 4 \). However, special cases arise as each critical point has an even power. No sign change occurs through these points, meaning the nature required for local extrema isn't met. The function's slope doesn't turn negative-to-positive or positive-to-negative at these points, so local minima or maxima are absent in the region of these critical points.

Concavity

Concavity is a measure of how a function curves. When a function is concave up, it resembles a smile, and when concave down, it looks like a frown. The sign of the second derivative \( f''(x) \) indicates concavity:

• \( f''(x) > 0 \): The function is concave up (think of a bowl).
• \( f''(x) < 0 \): The function is concave down (think of an upside-down bowl).

In this exercise, although we focus on the first derivative, understanding which parts curve upwards or downward can provide insights into the function's shape. Here, since the first derivative consists of even powers of polynomials, the behavior is often flat at critical points, with no switch in concave regions, reinforcing the absence of local extrema.

Sign Analysis

Sign analysis involves investigating the sign changes of \( f'(x) \) around critical points to determine increasing or decreasing behavior. It is vital in confirming the existence of local extrema, as changes from positive to negative indicate maxima, and changes from negative to positive suggest minima.
The derivative given is \( f'(x) = (x-1)^2 (x-2)^2 (x-3)^2 (x-4)^2 \), meaning:

• All terms are raised to an even power, resulting in non-negative outputs for any \( x \).
• No sign change occurs at the critical points \( x = 1, 2, 3, 4 \).

This leads us to deduce that the function maintains a constant direction at these points. The curves don’t have sections crossing above and below the x-axis around these critical points, demonstrating a lack of local extrema, as seen in this exercise.