Problem 24 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ f(x)=4-x^{3}, \quad-2 \leq x \leq 1 $$

Step-by-Step Solution

Answer

Absolute maximum is 12 at \(x = -2\), and minimum is 3 at \(x = 1\).

Step 1: Identify Critical Points

First, take the derivative of the function to find the critical points. The function is given by \( f(x) = 4 - x^3 \). The derivative is \( f'(x) = -3x^2 \). Set the derivative equal to zero to find critical points: \[ -3x^2 = 0 \] Solving for \( x \), we get \[ x = 0 \].

Step 2: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical point and at the endpoints of the interval. The endpoints given are \( x = -2 \) and \( x = 1 \). - For \( x = -2 \): \[ f(-2) = 4 - (-2)^3 = 4 + 8 = 12 \] - For \( x = 0 \): \[ f(0) = 4 - 0^3 = 4 \] - For \( x = 1 \): \[ f(1) = 4 - 1^3 = 3 \].

Step 3: Determine Absolute Maximum and Minimum Values

Compare the values obtained from the previous step to find the absolute maximum and minimum values. - The absolute maximum is \( f(-2) = 12 \). - The absolute minimum is \( f(1) = 3 \). Thus, the absolute maximum and minimum occur at \( x = -2 \) and \( x = 1 \) respectively.

Step 4: Graph the Function

Graph \( f(x) = 4 - x^3 \) over the interval \(-2 \leq x \leq 1 \). Plot the points \((-2, 12)\) and \((1, 3)\) on the graph, which represent the absolute maximum and minimum respectively. Also, plot the point \((0, 4)\) where the critical point occurs.

Step 5: Verify Coordinates on Graph

Check the plotted graph to ensure the coordinates match the calculated maximum and minimum points. The points are: - Absolute maximum at \((-2, 12)\). - Absolute minimum at \((1, 3)\).

Key Concepts

Critical Points

Critical points are where the function's derivative is zero or undefined. They're crucial because they indicate potential maximum or minimum points of the function. Finding these points helps us identify where the graph might change its direction or rate of increase or decrease.

For the function given in the exercise, there was only one critical point found at \( x = 0 \). This was determined by setting the derivative \(-3x^2\) to zero and solving for \(x\). Critical points might not always be where the absolute extrema are, but they are essential in locating potential extrema.

Absolute Extrema

Absolute extrema are the highest or lowest values that a function can achieve over a specific interval. In simpler terms, it's about figuring out the tallest peak or the deepest valley on a portion of a graph.

From the exercise, we found the absolute maximum value of the function \( f(x) = 4 - x^3 \) to be 12 at \( x = -2 \), and the absolute minimum value to be 3 at \( x = 1 \). To find these, we evaluated the function not just at the critical point but also at the interval's endpoints. It's because extrema can often occur at the edges of the interval rather than in its interior.

Derivative

The derivative of a function gives us the slope of the tangent line at any point along the curve. It is incredibly useful in understanding how the function behaves, such as when it's increasing or decreasing.

In our example, the derivative was \( f'(x) = -3x^2 \). Since this derivative is always non-positive in the interval \(-2 \leq x \leq 1\), it indicates the function is non-increasing. Because of the derivative, we could easily find the critical points and subsequently use them in determining possible extrema for the function.

Graphing Functions

Graphing a function gives a visual picture of it over a certain range and helps us to better understand how the function behaves. It can show where the function reaches its peaks and valleys.

In this exercise, the function \(f(x) = 4 - x^3\) was graphed over the interval \(-2 \leq x \leq 1\). By graphing, we could clearly see the coordinates where the function attains its absolute maximum and minimum values, as well as verify the critical point. Plotting critical points such as \( (0, 4) \) alongside the extrema points \( (-2, 12) \) and \( (1, 3) \), helped confirm the analytical findings practically.