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Chapter 12: Problem 19

Find the exact location of all the relative and absolute extrema of each function. \(h(t)=2 t^{3}+3 t^{2}\) with domain \([-2,+\infty)\)

Short Answer

Expert verified
The function \(h(t)=2t^3+3t^2\) with domain \([-2,+\infty)\) has a relative maximum at \(t=-1\), relative minimum at \(t=0\), and an absolute minimum at \((-2,-4)\).

Step by step solution

01

Find the critical points

First, differentiating \(h(t)\) with respect to \(t\) to obtain the first derivative: \(h'(t) = \frac{d}{dt}(2t^3 + 3t^2)\) Using the power rule, we get: \(h'(t) = 6t^2 + 6t\) Now, set the first derivative equal to 0 to find the critical points: \(6t^2 + 6t = 0\)
02

Determine open intervals where the function is increasing or decreasing

First, we will factor out \(6t\) from the equation found in Step 1: \(6t(t + 1) = 0\) This gives us two critical points: \(t = 0\) and \(t = -1\) Now, we need to determine the intervals where the function is increasing or decreasing. To do this, we can use a number line with partition points at our critical points and analyze the sign of \(h'(t)\) between the points: 1. Interval \((-\infty, -1)\): Choose a test point, say \(t = -2\), then \(h'(-2) = 6(-2)(-2 + 1) = 12 > 0\). So, the function is increasing in this interval. 2. Interval \((-1, 0)\): Choose a test point, say \(t = -\frac{1}{2}\), then \(h'(-\frac{1}{2}) = 6(-\frac{1}{2})(-\frac{1}{2} + 1) = -\frac{3}{2} < 0\). So, the function is decreasing in this interval. 3. Interval \((0, +\infty)\): Choose a test point, say \(t = 1\), then \(h'(1) = 6(1)(1+1) = 12 > 0\). The function is increasing in this interval.
03

Analyze the critical points and endpoints to classify them as relative extrema

From our analysis in Step 2, we have: 1. \(t = -1\): The function changes from increasing to decreasing, making this a relative maximum. 2. \(t = 0\): The function changes from decreasing to increasing, making this a relative minimum.
04

Find the absolute extrema (if any) by comparing the values at critical points and endpoints

Since the domain is \([-2,+\infty)\), the function has only one endpoint at -2. We should also consider the critical points previously obtained. Thus, we will compare the values of the function at these points: 1. \(h(-2) = 2(-2)^3 + 3(-2)^2 = -16 + 12 = -4\) 2. \(h(-1) = 2(-1)^3 + 3(-1)^2 = -2 + 3 = 1\) 3. \(h(0) = 2(0)^3 + 3(0)^2 = 0\) Since there's no upper bound for the function, there's no absolute maximum. As for the absolute minimum, the lowest value corresponds to the endpoint \(t = -2\) with a value of \(h(-2) = -4\). Thus, the absolute minimum occurs at \((-2,-4)\). In conclusion, the function \(h(t)=2t^3+3t^2\) has a relative maximum at \(t=-1\), a relative minimum at \(t=0\), and an absolute minimum at \((-2,-4)\) in the domain \([-2,+\infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
To understand the idea of critical points, we need to first get the first derivative of our function. In this case, the function is given as \( h(t) = 2t^3 + 3t^2 \), so the derivative is calculated as \( h'(t) = 6t^2 + 6t \).
Next, to locate the critical points, we set this derivative equal to zero: \( 6t^2 + 6t = 0 \). Solving this equation will help us find that the function has critical points at \( t = 0 \) and \( t = -1 \).
These critical points indicate where the function might change its behavior - shifting from increasing to decreasing, or vice versa. In simpler terms, they're spots where the function could have peaks or valleys.
Increasing and Decreasing Intervals
Analyzing where a function increases or decreases is essential for understanding its full behavior. To find these intervals for our function, we consider the sign of the derivative \( h'(t) \) between our critical points.
First, test the interval \((-\infty, -1)\) by choosing \( t = -2 \): \( h'(-2) = 12 \) indicates the function is increasing here.
Next, test the interval \((-1, 0)\) with \( t = -\frac{1}{2} \): \( h'(-\frac{1}{2}) = -\frac{3}{2} \) shows it's decreasing in this interval.
Finally, test \((0, +\infty)\) using \( t = 1 \): \( h'(1) = 12 \) confirms the function is again increasing.
  • Interval \((-\infty, -1)\): Increasing
  • Interval \((-1, 0)\): Decreasing
  • Interval \((0, +\infty)\): Increasing
Absolute Minimum
Identifying the absolute minimum involves evaluating the function at our critical and endpoint values and comparing them to find the smallest value. Here, the domain of our function is restricted to \([-2, +\infty)\).
For \( h(t) \):
  • \( h(-2) = -4 \)
  • \( h(-1) = 1 \)
  • \( h(0) = 0 \)

The lowest value of these is \(-4\), found at \( t = -2 \). Hence, the function reaches its absolute minimum at this endpoint, giving us the point \((-2, -4)\). This value signifies the lowest point within the domain where the function can reach, despite it continuing to increase toward infinity in the future.

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Most popular questions from this chapter

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