Problem 40 - Thomas Calculus

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur.$$ k(x)=x^{2 / 3}\left(x^{2}-4\right) $$

Step-by-Step Solution

Answer

Increasing on (-1,0) and (1,inf); decreasing on (-inf,-1) and (0,1). Local min at x=+/-1, k=-3; local max at x=0, k=0.

Step 1: Find the Derivative

\(k(x) = x^{2/3}(x^2-4)\)
\(k'(x) = \frac{2}{3}x^{-1/3}(x^2-4) + x^{2/3}(2x) = \frac{2(x^2-4)}{3x^{1/3}} + 2x^{5/3}\)
\(= \frac{2(x^2-4) + 6x^2}{3x^{1/3}} = \frac{8x^2 - 8}{3x^{1/3}} = \frac{8(x^2-1)}{3x^{1/3}}\)

Step 2: Find Critical Points

\(k'(x) = 0\) when \(x^2 - 1 = 0\), so \(x = \pm 1\). Also \(k'\) is undefined at \(x = 0\).
Critical points: \(x = -1, 0, 1\).

Step 3: Determine Intervals

Test signs of \(k'(x)\):
\(x < -1\): numerator \(+\), denominator \(-\) → \(k' < 0\) (decreasing)
\(-1 < x < 0\): numerator \(-\), denominator \(-\) → \(k' > 0\) (increasing)
\(0 < x < 1\): numerator \(-\), denominator \(+\) → \(k' < 0\) (decreasing)
\(x > 1\): numerator \(+\), denominator \(+\) → \(k' > 0\) (increasing)

Step 4: Extrema

Local min at \(x = -1\): \(k(-1) = 1 \cdot (-3) = -3\)
Local max at \(x = 0\): \(k(0) = 0\)
Local min at \(x = 1\): \(k(1) = 1 \cdot (-3) = -3\)
No absolute extremes (function → \(\infty\) as \(x \to \pm\infty\)).

Key Concepts

Derivative

The derivative is a crucial concept in calculus, representing the rate at which a function is changing at any given point. When we talk about finding the derivative of a function like \(k(x) = x^{2/3}(x^2 - 4)\), we aim to understand how small changes in \(x\) lead to changes in \(k(x)\). This is essential for determining whether a function is increasing or decreasing.

To find the derivative, we often use rules such as the product rule, chain rule, and power rule. The product rule is especially useful when dealing with functions that are products of two or more simpler functions. As seen in the exercise, if \(u(x) = x^{2/3}\) and \(v(x) = x^2 - 4\), then the product rule tells us that:

• \(k'(x) = u'(x)v(x) + u(x)v'(x)\)

This formula helps us decompose the task of differentiation into manageable parts.

Breaking it down, finding \(u'(x) = \frac{2}{3}x^{-1/3}\) and \(v'(x) = 2x\) involves using the power rule, which is a shortcut for differentiating functions that involve powers of \(x\). The derivative \(k'(x)\) thus gives us a new function that we can analyze to understand the behavior of the original function \(k(x)\).

Increasing and Decreasing Intervals

Identifying increasing and decreasing intervals of a function involves analyzing its derivative \(k'(x)\). The behavior of \(k'(x)\) determines whether the original function \(k(x)\) is rising or falling over specified intervals.

Here's the essential idea:

• If \(k'(x) > 0\) for a particular interval, the function is increasing over that interval.
• If \(k'(x) < 0\), the function is decreasing.
• If \(k'(x) = 0\), we have a potential turning point that may indicate a local maximum or minimum.

Checking where \(k'(x)\) equals zero helps us find critical points—places where the function might change from increasing to decreasing or vice versa. After solving \(k'(x) = 0\), the roots or solutions you get are the critical points of the function. By testing the sign of \(k'(x)\) before and after these critical points, you confirm the nature of the intervals (increasing or decreasing). This step is pivotal for sketching the graph and understanding the function's overall behavior.

Extreme Values

Extreme values in calculus refer to the maximum or minimum points on a function's curve. These points can be local (only higher or lower within a given region) or absolute/global (highest or lowest over the entire range of the function).

To find these values for a function like \(k(x) = x^{2/3}(x^2 - 4)\), we utilize critical points discovered during our analysis of increasing and decreasing intervals. Critical points occur where the derivative \(k'(x)\) is zero or undefined.

• A local maximum is a high point surrounded by lower points, while a local minimum is a low point surrounded by higher ones.
• An absolute maximum is the highest point on the entire function, and an absolute minimum is the lowest.

To identify these extremes:

• Analyze the sign of \(k'(x)\) around the critical points.
• Check the endpoints and use the first or second derivative test to confirm if a point is a maximum or minimum.

This analysis allows us to understand not just where extreme values occur, but also their nature—helping us answer questions about the largest or smallest values the function can achieve.