Problem 30 - University Calculus: Early Transcendentals

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. $$K(t)=15 t^{3}-t^{5}$$

Step-by-Step Solution

Answer

Function increases on \((-\infty, -3)\), \((-3, 0)\), \((0, 3)\); decreases on \((3, \infty)\). Local max of 27 at \(t=3\), local min of -27 at \(t=-3\).

Step 1: Find the derivative of the function

To determine where the function is increasing or decreasing, we first need to find its derivative. The derivative of the function \( K(t) = 15t^3 - t^5 \) is computed using power rule:\[ K'(t) = \frac{d}{dt}(15t^3) - \frac{d}{dt}(t^5) = 45t^2 - 5t^4 \]

Step 2: Solve for critical points

Find the critical points by solving \( K'(t) = 0 \):\[ 45t^2 - 5t^4 = 0 \]Factor out common terms:\[ 5t^2(9 - t^2) = 0 \]So, this gives us:\[ t^2 = 0 \] or \[ 9 - t^2 = 0 \]Solving these, we find the critical points are \( t = 0, 3, -3 \).

Step 3: Determine intervals of increase and decrease

Using the critical points \( t = -3, 0, 3 \), test the sign of \( K'(t) \) in each interval:- For \( t < -3 \), choose \( t = -4 \): \( K'(-4) = 45(-4)^2 - 5(-4)^4 > 0 \) (function is increasing)- For \( -3 < t < 0 \), choose \( t = -1 \): \( K'(-1) = 45(-1)^2 - 5(-1)^4 > 0 \) (function is increasing)- For \( 0 < t < 3 \), choose \( t = 1 \): \( K'(1) = 45(1)^2 - 5(1)^4 > 0 \) (function is increasing)- For \( t > 3 \), choose \( t = 4 \): \( K'(4) = 45(4)^2 - 5(4)^4 < 0 \) (function is decreasing)Thus, the function is increasing on \((-\infty, -3) \), \((-3, 0)\), and \((0, 3)\) and decreasing on \((3, \infty)\).

Step 4: Identify local and absolute extreme values

Evaluate \( K(t) \) at the critical points to find extremums:- \( K(0) = 15(0)^3 - (0)^5 = 0 \)- \( K(3) = 15(3)^3 - (3)^5 = 270 - 243 = 27 \)- \( K(-3) = 15(-3)^3 - (-3)^5 = -270 + 243 = -27 \)Since the function changes from increasing to decreasing at \( t = 3 \), there is a local maximum at \( t = 3 \) with the value 27. Since the behavior is decreasing for \( t > 3 \), \( t = -3 \), \( K(-3) = -27 \) could appear as a local minimum, but since the function is only specified over the range of all real numbers, there's no symmetry constraint here that allows identification of global extrema solely by point testing.

Key Concepts

Derivative

A derivative in calculus is like a mathematical tool that helps us understand how a function changes. Imagine it as a speedometer that measures how fast something is moving, but instead of speed, it measures how quickly a function's value is changing with respect to its input.

To find the derivative of a function, we use rules and formulas from calculus, one of the most common being the "power rule." In the example of the function \( K(t) = 15t^3 - t^5 \), finding the derivative means applying the power rule to figure out a new function that describes the rate of change of \( K(t) \).

The power rule simply tells us that if you have a term like \( x^n \), its derivative will be \( nx^{n-1} \). So \( 15t^3 \) becomes \( 45t^2 \) and \( -t^5 \) becomes \( -5t^4 \). The derivative \( K'(t) = 45t^2 - 5t^4 \) is then used to gather insights about the original function's behavior.

Critical Points

Critical points are key locations on the graph of a function where the derivative is zero or undefined. These points are crucial because they usually signal where the function might change behavior, such as going from increasing to decreasing, which might indicate a peak or a valley.

To find these points, you need to solve the derivative equation \( K'(t) = 0 \). In our example, it led to the equation \( 45t^2 - 5t^4 = 0 \). By factoring it, we discover the critical points \( t = 0, 3, -3 \), indicating potential places where the function's slope changes direction.

Identifying critical points is often the first step in analyzing a function's overall behavior, so they hold a lot of importance when investigating curves.

Increasing and Decreasing Intervals

Once the derivative is found, we use it to look at where the function is increasing or decreasing. This step involves analyzing the sign of the derivative over different intervals.

If the derivative \( K'(t) \) is positive on an interval, the function is increasing there. This means as the input \( t \) goes up, so does the output \( K(t) \). Conversely, if \( K'(t) \) is negative, then the function is decreasing, indicating that as \( t \) increases, \( K(t) \) decreases.

In our case, the function was shown to increase on \((-fty, -3)\), \((-3, 0)\), and \((0, 3)\), and then decrease on \((3, fty)\). This pattern reveals the overall zig-zag path of the graph, dictating where the peaks and troughs might be on the function's landscape.

Local and Absolute Extrema

The terms local and absolute extrema refer to the highest or lowest points on a graph. **Local extrema** are peaks or valleys within a limited section of the graph. Think of them as hills and dips in the geographic terrain. Meanwhile, **absolute extrema** refer to the highest or lowest points over the entire graph, the "highest mountain" or the "deepest valley."

To find these extrema, we evaluate the function at critical points and also check the intervals where the function is defined. In the provided example, at \( t = 3 \), the function goes from increasing to decreasing, creating a local maximum at \( K(3) = 27 \). At \( t = -3 \), the opposite happens, leading to a local minimum at \( K(-3) = -27 \). Local extrema provide insights into the function's relative performance, while absolute extrema define the overall constraints or bounds.