Problem 31 - 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

\( f(t) \) increases on \((0, 2)\) and \((4, 5)\), decreases on \((2, 4)\); maxima at \( t = 2 \), minima at \( t = 4 \).

Step 1: Identify Critical Points

To determine where the function \( f(t) \) transitions from increasing to decreasing or vice versa, find the critical points by setting the derivative equal to zero: \[ f^{\prime}(t) = t^3 - 6t^2 + 8t = 0. \] Factor to solve for \( t \).

Step 2: Solve for Critical Points

Factor the equation in step 1: \[ t(t^2 - 6t + 8) = 0. \] This gives \( t = 0 \) or \( t^2 - 6t + 8 = 0.\) Solving the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get \( t = 2 \) and \( t = 4 \). Thus, critical points are \( t = 0, 2, 4 \).

Step 3: Test Intervals Around Critical Points

To determine interval behavior, select test points from the intervals \((0, 2)\), \((2, 4)\), and \((4, 5)\). Evaluate \( f^{\prime}(t) \) at each interval point:- For \( t = 1 \), \( f^{\prime}(1) > 0 \), so \( f(t) \) is increasing.- For \( t = 3 \), \( f^{\prime}(3) < 0 \), so \( f(t) \) is decreasing.- For \( t = 4.5 \), \( f^{\prime}(4.5) > 0 \), so \( f(t) \) is increasing.

Step 4: Determine Local Maximum and Minimum

Using the information from Step 3, find where \( f(t) \) has local maxima and minima:- At \( t = 2 \), \( f(t) \) changes from increasing to decreasing, indicating a local maximum.- At \( t = 4 \), \( f(t) \) changes from decreasing to increasing, indicating a local minimum.

Key Concepts

Critical Points

Critical points are essential in understanding how a function behaves within a specific interval. These points occur where the derivative of a function is equal to zero or where the derivative is undefined. For our function, the derivative is given by \(f^{\prime}(t) = t^{3} - 6t^{2} + 8t\). To find the critical points, we set \(f^{\prime}(t) = 0\) and solve for \(t\). Here’s how this process unfolds for our specific derivative:

• First, factor the equation: \(t(t^{2} - 6t + 8) = 0\).
• This gives one solution as \(t = 0\) because of the factor \(t\) itself.
• Next, solve the remaining quadratic equation \(t^{2} - 6t + 8 = 0\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).
• This results in additional critical points at \(t = 2\) and \(t = 4\).

These critical points \(t = 0, 2,\) and \(4\) help us explore how and where the function transitions between increasing and decreasing behavior.

Increasing and Decreasing Functions

Increasing and decreasing functions give insight into the behavior of \(f(t)\) across specific intervals by indicating where the function is rising or falling. This is determined by analyzing the derivative's sign around the critical points. A positive derivative means the function is increasing, while a negative derivative means it is decreasing.Here’s what we find in our exercise:

• From \(t = 0\) to \(t = 2\), test point \(t = 1\) shows \(f^{\prime}(1) > 0\), so the function is increasing.
• From \(t = 2\) to \(t = 4\), test point \(t = 3\) reveals \(f^{\prime}(3) < 0\), indicating that the function is decreasing.
• From \(t = 4\) to \(t = 5\), test point \(t = 4.5\) results in \(f^{\prime}(4.5) > 0\), showing the function is increasing again.

By evaluating small test numbers within these intervals, we can determine the exact nature of change within \(f(t)\), leading us to identify key patterns in its behavior over the interval [0, 5].

Local Maximum and Minimum

Local maximum and minimum points are special types of critical points that represent the highest or lowest values in the surrounding interval of the function. These are identified by observing changes in the sign of the derivative in surrounding intervals.In this exercise, we notice the following:

• At \(t = 2\), the function transitions from increasing ( extit{from the left}) to decreasing ( extit{to the right}), marking a local maximum.
• Conversely, at \(t = 4\), the transition is from decreasing to increasing, which characterizes a local minimum.

Identifying these points helps give a snapshot of where a function reaches peak or valley values locally, thus illustrating significant aspects of \(f(t)\)'s overall shape and distribution over the specified range.