Problem 29 - 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. $$ H(t)=\frac{3}{2} t^{4}-t^{6} $$

Step-by-Step Solution

Answer

H is increasing on (-1, 0) and (0, 1), decreasing on (-∞, -1) and (1, ∞). Local minima at t = ±1, local maximum at t = 0.

Step 1: Find the derivative

First, we need to find the derivative of the function \( H(t) = \frac{3}{2} t^{4} - t^{6} \). To get this, apply the power rule to each term:The derivative of \( \frac{3}{2} t^4 \) is \( 4 \cdot \frac{3}{2} \cdot t^{3} = 6t^3 \).The derivative of \( -t^6 \) is \( -6t^5 \).Thus, the derivative is \( H'(t) = 6t^3 - 6t^5 \).

Step 2: Factor the derivative

To simplify our analysis, factor the derivative.\[H'(t) = 6t^3(1 - t^2)\]This will help us find the critical points where \( H'(t) = 0 \).

Step 3: Find critical points

Set \( H'(t) = 0 \) to find critical points:\[6t^3(1 - t^2) = 0\]This gives us two equations:1. \( 6t^3 = 0 \) leading to \( t=0 \).2. \( 1 - t^2 = 0 \) leading to \( t^2 = 1 \), thus \( t = \pm 1 \).The critical points are \( t = 0, 1, -1 \).

Step 4: Determine intervals of increase/decrease

Using the critical points, set up the intervals: 1. \( (-\infty, -1) \)2. \( (-1, 0) \)3. \( (0, 1) \)4. \( (1, \infty) \)Choose a test point in each interval and plug it into \( H'(t) \) to determine if \( H(t) \) is increasing or decreasing:- For \((-\infty, -1)\), test \( t = -2 \): \( H'(-2) = 6(-2)^3(1-(-2)^2) = -96 \), decreasing.- For \((-1, 0)\), test \( t = -0.5 \): \( H'(-0.5) = 6(-0.5)^3(1-(-0.5)^2) = 1.125 \), increasing.- For \((0, 1)\), test \( t = 0.5 \): \( H'(0.5) = 6(0.5)^3(1-(0.5)^2) = 1.125 \), increasing.- For \((1, \infty)\), test \( t = 2 \): \( H'(2) = 6(2)^3(1-(2)^2) = -96 \), decreasing.Thus, \( H(t) \) is increasing on \((-1, 0) \cup (0, 1)\) and decreasing on \((-fty, -1) \cup (1, \infty)\).

Step 5: Identify extreme values

Evaluate \( H(t) \) at the critical points to identify potential local or absolute extrema:- At \( t = -1 \), \( H(-1) = \frac{3}{2}(-1)^4 - (-1)^6 = \frac{3}{2} - 1 = \frac{1}{2} \).- At \( t = 0 \), \( H(0) = \frac{3}{2}(0)^4 - (0)^6 = 0 \).- At \( t = 1 \), \( H(1) = \frac{3}{2}(1)^4 - (1)^6 = \frac{3}{2} - 1 = \frac{1}{2} \).Since \( H(t) \) is decreasing to the left of \( t = -1 \) and increasing for \( t \) in \((-1, 0) \), \( H(-1) = \frac{1}{2} \) is a local minimum.Since \( H(t) \) is increasing to \( t = 0 \) and then decreasing for \( t \) in \((0, 1) \), \( H(0) = 0 \) is a local maximum.Since \( H(t) \) is decreasing to the right of \( t = 1 \), \( H(1) = \frac{1}{2} \) is a local minimum.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus that measures the rate at which a function changes. In simple terms, it tells us how the function is behaving at any point on its curve. It represents the slope of the tangent line to the function at a particular point. For the function \( H(t) = \frac{3}{2} t^{4} - t^{6} \), we find the derivative using the power rule. This derivative shows us the behavior of the function, helping us to pinpoint where it increases or decreases. By differentiating terms like \( \frac{3}{2} t^4 \) and \( -t^6 \), we can discern the rate of change for each component of the function. Hence, the derivative of \( H(t) \) is \( 6t^3 - 6t^5 \). This is crucial for identifying critical points and understanding the function's overall dynamics.

Critical Points

Critical points are those values of \( t \) where the derivative is either zero or undefined. At these points, the function may have a peak, valley, or saddle point. For the function \( H(t) \), we find the critical points by setting the derivative \( H'(t) = 6t^3 - 6t^5 \) to zero and solving. Factoring gives \( H'(t) = 6t^3(1 - t^2) = 0 \), leading to \( t = 0, \pm 1 \). These points are essential as they indicate where the function changes direction, meaning the behavior of the function may change from increasing to decreasing, or vice versa.

Increasing and Decreasing Intervals

To understand where the function is increasing or decreasing, we evaluate the sign of the derivative \( H'(t) \) in the intervals defined by the critical points. The derivative guides whether the function's slope is positive or negative. If \( H'(t) > 0 \), the function is increasing; if \( H'(t) < 0 \), it is decreasing. For instance, testing points in intervals such as \((-\infty, -1), (-1, 0), (0, 1), (1, \infty)\) reveals that \( H(t) \) is increasing between (-1, 0), (0, 1) and decreasing at (-\infty, -1), (1, \infty). This analysis helps predict how the function behaves over different sections of its domain.

Local Maximum and Minimum

Local maxima and minima are the highest or lowest points in a function's immediate vicinity. They occur at critical points where the function changes direction. For \( H(t) \), we determine these by analyzing the function's behavior around the critical points \( t = -1, 0, 1 \). The function is decreasing before \( t = -1 \) and increases afterward, making \( H(-1) = \frac{1}{2} \) a local minimum. Conversely, \( t = 0 \) experiences a change from increasing to decreasing, resulting in a local maximum at \( H(0) = 0 \). At \( t = 1 \), the same pattern as \( t = -1 \) occurs, thus \( H(1) = \frac{1}{2} \) is also a local minimum. Understanding these points gives insight into the peaks and troughs of the function.

Power Rule

The power rule is a shortcut for finding the derivative of polynomial terms. It states that for any function \( f(x) = ax^n \), its derivative is \( f'(x) = nax^{n-1} \). This rule simplifies differentiation by allowing quick computation of individual terms. In our exercise, it efficiently transforms terms like \( \frac{3}{2} t^4 \) into \( 6t^3 \), and \( -t^6 \) into \( -6t^5 \). Applying the power rule helps reduce complex terms and prepare them for further analysis, such as finding critical points or determining intervals of increase and decrease. It's an essential tool for calculus students to master, as it lays the groundwork for understanding derivatives of more complex functions.