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

Short Answer

Expert verified
The function is increasing on \((-\infty, -1)\) and \((0, 1)\), decreasing on \((-1, 0)\) and \((1, \infty)\). Local maxima at \(t = -1, 1\) and local minimum at \(t = 0\); no absolute extrema.

Step by step solution

01

Find the Derivative

To find the intervals where the function is increasing or decreasing, we first need to find the derivative of the function. The function given is \(H(t) = \frac{3}{2} t^4 - t^6\). The derivative, \(H'(t)\), is found using the power rule for derivatives, where \(\frac{d}{dt} t^n = n t^{n-1}\). Therefore, \(H'(t) = \frac{d}{dt}(\frac{3}{2}t^4) - \frac{d}{dt}(t^6)\). Calculating this, we get \(H'(t) = 6t^3 - 6t^5\).
02

Factor the Derivative

To find critical points, let's factor the derivative \(H'(t) = 6t^3 - 6t^5\). Factoring out the common term \(6t^3\), we have \(H'(t) = 6t^3(1 - t^2)\).
03

Solve for Critical Points

Set \(H'(t) = 0\) to find the critical points. Thus, \(6t^3(1-t^2) = 0\). This equation is satisfied if \(6t^3 = 0\) or \(1 - t^2 = 0\). Solving these, we get \(t = 0\) from \(t^3 = 0\), and \(t = \pm1\) from \(t^2 = 1\). Thus, critical points are \( t = -1, 0, 1 \).
04

Determine Intervals of Increase/Decrease

Next, test the intervals defined by the critical points \(-\infty, -1, 0, 1, \infty\) using \(H'(t)\):1. For \(t < -1\), choose \(t = -2\), then \(H'(-2) > 0\).2. For \(-1 < t < 0\), choose \(t = -0.5\), then \(H'(-0.5) < 0\).3. For \(0 < t < 1\), choose \(t = 0.5\), then \(H'(0.5) > 0\).4. For \(t > 1\), choose \(t = 2\), then \(H'(2) < 0\).This analysis tells us that \(H(t)\) is increasing on \((-\infty, -1)\) and \((0, 1)\), and decreasing on \((-1, 0)\) and \((1, \infty)\).
05

Identify Local and Absolute Extreme Values

Check the derivative changes at \(t = -1, 0, 1\):- At \(t = -1\), \(H'(t)\) changes from positive to negative, indicating a local maximum.- At \(t = 0\), \(H'(t)\) changes from negative to positive, indicating a local minimum.- At \(t = 1\), \(H'(t)\) changes from positive to negative, indicating another local maximum.Calculate the function values:- \(H(-1) = \frac{3}{2}(-1)^4 - (-1)^6 = \frac{3}{2} - 1 = \frac{1}{2}\).- \(H(0) = \frac{3}{2}(0)^4 - (0)^6 = 0\).- \(H(1) = \frac{3}{2}(1)^4 - (1)^6 = \frac{3}{2} - 1 = \frac{1}{2}\).So, there are local maxima of \(\frac{1}{2}\) at \(t = -1\) and \(t = 1\), and a local minimum of \(0\) at \(t = 0\). The function does not have absolute extrema because it tends towards \(-\infty\) as \(t\) approaches \(+\infty\) or \(-\infty\).

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Key Concepts

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

Derivative
A derivative is a mathematical tool that helps us understand how a function behaves, particularly how it changes. In simpler terms, the derivative of a function at a point gives us the slope of the function at that point. It tells us whether the function is going up or down and how steeply it's doing so. For the function \(H(t) = \frac{3}{2} t^4 - t^6\), we use the power rule to find its derivative: \(H'(t) = 6t^3 - 6t^5\). The power rule states that the derivative of \(t^n\) is \(nt^{n-1}\). This makes it easier to find the rate of change of polynomial functions. By factoring \(6t^3\) from \(H'(t) = 6t^3 - 6t^5\), we get \(H'(t) = 6t^3(1 - t^2)\). This step is crucial because it sets the stage for finding critical points, which are where the behavior of the function transitions.
Increasing and Decreasing Intervals
The intervals where a function is increasing or decreasing depend on the sign of its derivative. When the derivative is positive, the function is rising, indicating an increasing interval. Conversely, a negative derivative means the function is falling, showing a decreasing interval. For our function \(H(t)\), the critical points found from \(H'(t) = 0\) are \(t = -1, 0, 1\). By testing values in these intervals, we determined:
  • The function is increasing on \((-\infty, -1)\) and \((0, 1)\)
  • The function is decreasing on \((-1, 0)\) and \((1, \infty)\)
This detailed analysis of intervals provides insights into where the function gains or loses values as \(t\) changes.
Local Extreme Values
Local extreme values are valuable insights that tell us about peaks and troughs in a function. Local maxima are points where the function reaches a peak, and cannot increase further without first decreasing. Local minima are points where the function hits a low, and cannot decrease without first increasing. Using the sign changes in the derivative \(H'(t)\), we find local extrema for \(H(t)\):
  • Local Maxima at \(t = -1\) and \(t = 1\) with function value \(H(t) = \frac{1}{2}\)
  • Local Minimum at \(t = 0\) with function value \(H(t) = 0\)
Unlike absolute extrema, which are the highest or lowest values of a function over its entire range, local extrema are specific to smaller intervals. For this function \(H(t)\), there are no absolute extrema as it trends towards \(-\infty\) beyond these turning points. Understanding these concepts is key to mastering function analysis.

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