Problem 18 - Applied Calculus

Question

The derivative of \(f(t)\) is given by \(f^{\prime}(t)=t^{3}-6 t^{2}+8 t\) for \(0 \leq t \leq 5\). Graph \(f^{\prime}(t)\), and describe how the function \(f(t)\) changes over the interval \(t=0\) to \(t=5 .\) When is \(f(t)\) increasing and when is it decreasing? Where does \(f(t)\) have a local maximum and where does it have a local minimum?

Step-by-Step Solution

Answer

\(f(t)\) is increasing on \(0 < t < 2\) and \(4 < t < 5\); decreasing on \(2 < t < 4\). Local max at \(t = 2\) and local min at \(t = 4\).

Step 1: Understand Derivative and Critical Points

First, we need to find the critical points of the function by setting the derivative equal to zero: \[ t^3 - 6t^2 + 8t = 0. \] This equation can be factored as: \[ t(t^2 - 6t + 8) = 0. \]Further factoring yields:\[ t(t-2)(t-4) = 0. \]Thus, the critical points are \( t = 0, 2, \) and \( 4. \)

Step 2: Determine Intervals of Increase and Decrease

To know where \( f(t) \) is increasing or decreasing, we check the sign of \( f'(t) \) between the critical points.- For \( 0 < t < 2 \), choose \( t = 1 \): \[ f'(1) = 1^3 - 6 imes 1^2 + 8 imes 1 = 3 \] (positive, so increasing).- For \( 2 < t < 4 \), choose \( t = 3 \): \[ f'(3) = 3^3 - 6 imes 3^2 + 8 imes 3 = -3 \] (negative, so decreasing).- For \( 4 < t < 5 \), choose \( t = 4.5 \): \[ f'(4.5) = (4.5)^3 - 6 imes (4.5)^2 + 8 imes 4.5 = 2.625 \] (positive, so increasing).

Step 3: Identify Local Maxima and Minima

From the sign changes:- At \( t = 2 \), \( f'(t) \) changes from positive to negative, indicating a local maximum.- At \( t = 4 \), \( f'(t) \) changes from negative to positive, indicating a local minimum.

Step 4: Sketch the Graph

Sketch or simply understand the graph of \( f'(t) = t^3 - 6t^2 + 8t \):- The x-axis intercepts at points \( t = 0, 2, \) and \( 4 \).- Increasing on intervals \( (0, 2) \) and \( (4, 5) \).- Decreasing on interval \( (2, 4) \).This helps to visually confirm where the function \( f(t) \) is increasing, decreasing, and has local extrema.

Key Concepts

Critical Points

To find critical points, we set the derivative of the function equal to zero. These points occur where there could be a peak, valley, or a flat spot on the graph. For the function with derivative \(f^{\prime}(t) = t^{3}-6t^{2}+8t\), we start by solving \(t^{3} - 6t^{2} + 8t = 0\). This equation can be factored as \(t(t-2)(t-4) = 0\), giving us critical points at \(t = 0, 2, \text{ and } 4\). Critical points are crucial because they help us determine the behavior of the function. Each critical point is a candidate for a local maximum or minimum, and understanding them aids in analyzing the function's overall dynamics.

Intervals of Increase and Decrease

Once we've identified the critical points, we check the intervals between them to see if the function is increasing or decreasing.- For \(0 < t < 2\), we test a point within this interval like \(t = 1\). Since \(f'(1) > 0\), the function is increasing.- For \(2 < t < 4\), we choose \(t = 3\) and find that \(f'(3) < 0\), so here the function is decreasing.- Lastly, for \(4 < t < 5\), testing \(t = 4.5\) shows \(f'(4.5) > 0\), indicating the function is increasing again.Understand that the sign of the derivative \(f'(t)\) tells us if the original function \(f(t)\) is rising or falling in these segments. Positive derivative means increasing, while negative means decreasing.

Local Maximum and Minimum

Identifying local maxima and minima involves looking at the transitions of \(f'(t)\) between positive and negative signs. - At \(t = 2\), \(f'(t)\) changes from positive to negative. This change tells us \(f(t)\) reaches a peak here, signaling a local maximum.- Conversely, at \(t = 4\), \(f'(t)\) shifts from negative to positive, suggesting \(f(t)\) has a dip, which is a local minimum.These local extrema provide valuable insights into the function's graph. It's where the function behaves contrary to its nearby intervals and thus stands as significant markers to sketching or understanding the function's behavior over the given domain.