Problem 18 - Applied Calculus

Question

Find all critical points and then use the first-derivative test to determine local maxima and minima. Check your answer by graphing. $$f(x)=\left(x^{3}-8\right)^{4}$$

Step-by-Step Solution

Answer

Critical points are at \(x=0\) (local max) and \(x=2\) (local min).

Step 1: Find the First Derivative

To find the critical points, we first need the first derivative of the function. Given the function \[f(x) = (x^3 - 8)^4\]we use the chain rule to derive it. Let \(u = x^3 - 8\), then \(f(x) = u^4\). The derivative \( \frac{du}{dx} = 3x^2\) and \(\frac{df}{du} = 4u^3\). Apply the chain rule to get:\[\frac{df}{dx} = 4u^3 \cdot 3x^2 = 12x^2(x^3 - 8)^3\]

Step 2: Set the Derivative to Zero

Critical points occur where the derivative is zero or undefined. Set the first derivative \(12x^2(x^3 - 8)^3 = 0\). Since this is a product, set each factor to zero:\[12x^2 = 0\] and \[(x^3 - 8)^3 = 0\].For \(12x^2 = 0\), we get \(x = 0\). For \((x^3 - 8)^3 = 0\), we solve \(x^3 = 8\), yielding \(x = 2\). So the potential critical points are \(x = 0\) and \(x = 2\).

Step 3: Apply the First-Derivative Test

Use the first-derivative test around each critical point to determine the nature of critical points. For \(x = 0\):- Choose test points like \(x = -1\) and \(x = 1\). - At \(x = -1\), \(f'(x) = 12(-1)^2((-1)^3 - 8)^3 > 0\). - At \(x = 1\), \(f'(x) = 12(1)^2(1^3 - 8)^3 < 0\).Hence, \(x = 0\) is a local maximum.For \(x = 2\):- Choose test points like \(x = 1\) and \(x = 3\). - At \(x = 1\), \(f'(x) = 12(1)^2((1)^3 - 8)^3 < 0\). - At \(x = 3\), \(f'(x) = 12(3)^2((3)^3 - 8)^3 > 0\).Hence, \(x = 2\) is a local minimum.

Step 4: Verify by Graphing

Graph the function \(f(x) = (x^3 - 8)^4\) to visualize the critical points and confirm their nature. A graphing calculator or software shows that at \(x = 0\), the curve has a peak (local maximum), and at \(x = 2\), it has a trough (local minimum). This confirms the analysis done with the first-derivative test.

Key Concepts

Critical Points

Critical points are locations on a graph where the slope of the function is zero or where the derivative does not exist. Finding these points is crucial in understanding the overall behavior of the function, especially when seeking peaks (maximums) or valleys (minimums).

For our function, \(f(x) = (x^3 - 8)^4\), we calculate the first derivative and set it to zero to identify these critical locations. This involves us rewriting the function in terms of \(u = x^3 - 8\), differentiating with respect to \(x\), and solving for the points where the derivative either equals zero or becomes undefined.

In this particular problem, the derivative equation \(12x^2(x^3 - 8)^3 = 0\) gives us critical points at \(x = 0\) and \(x = 2\). These are the coordinates we'll examine further to understand the behavior of the function at these points.

First-Derivative Test

The first-derivative test is a method that helps us determine whether a critical point of a function is a local maximum or minimum. After finding the critical points, we select test points around these points and evaluate the derivative at these locations.

For \(x = 0\) and \(x = 2\), we check surrounding values to see how the slope of the graph changes direction:

• At \(x = 0\), we use test points like \(x = -1\) and \(x = 1\). If the derivative changes from positive to negative, it indicates a local maximum.
• At \(x = 2\), examining points like \(x = 1\) and \(x = 3\), if the derivative changes from negative to positive, it suggests a local minimum.

This test shows that \(x = 0\) is a local maximum (since the sign of the slope changes from positive to negative) and \(x = 2\) is a local minimum (as the slope moves from negative to positive).

Local Maxima and Minima

Local maxima and minima refer to the highest and lowest points in the immediate vicinity on a graph, respectively. These are critical for understanding the function's behavior and can help analyze how the function grows or shrinks.

In our analysis of \(f(x) = (x^3 - 8)^4\), we found the following:

• The local maximum at \(x = 0\) means the function reaches a peak point there. This is typically characterized by the curve being higher than nearby points.
• The local minimum at \(x = 2\) indicates a trough, where the function dips lower before rising again.

The nature of these points was confirmed by the first-derivative test, which helps give a clear picture of how the function behaves around these critical areas.

Graphing Functions

Graphing functions provides a visual interpretation of the mathematical behavior of the equation. Seeing how the function behaves at critical points helps verify our earlier analytical findings.

By plotting \(f(x) = (x^3 - 8)^4\), we can see:

• At \(x = 0\), the graph has a noticeable peak, corroborating our determination of a local maximum.
• At \(x = 2\), it displays a smooth dip, confirming it as a local minimum.

Graphing isn't just useful for verification but also for understanding concepts intuitively. Tools like graphing calculators or software can make these visualizations accessible, allowing students to see the functions' stories unfold, enhancing their grasp of calculus concepts like critical points, maxima, and minima.