Problem 21 - Thomas Calculus

Question

In Exercises \(15-30\) , 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. $$ h(x)=\sqrt[3]{x}, \quad-1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

Absolute maximum: 2 at \(x=8\), absolute minimum: -1 at \(x=-1\).

Step 1: Identify Critical Points

To find critical points of the function, we need to calculate the derivative of \(h(x)\). Given \(h(x) = \sqrt[3]{x} = x^{1/3}\), we find the derivative:\[ h'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\]The derivative is undefined at \(x=0\). Hence, \(x=0\) is a critical point.

Step 2: Evaluate Function at Endpoints and Critical Points

Evaluate \(h(x)\) at the critical point and interval endpoints. We calculate:- \(h(-1) = \sqrt[3]{-1} = -1\)- \(h(0) = \sqrt[3]{0} = 0\)- \(h(8) = \sqrt[3]{8} = 2\)These give us the values at the critical point and endpoints.

Step 3: Determine Absolute Extrema

Examining the values from Step 2, we compare: - For \(h(-1) = -1\) - For \(h(0) = 0\) - For \(h(8) = 2\)Thus, the absolute maximum value is \(2\) at \(x = 8\), and the absolute minimum value is \(-1\) at \(x = -1\).

Step 4: Graph the Function and Mark Extrema

Plot the function \(h(x) = \sqrt[3]{x}\) on the interval \([-1, 8]\). Highlight the points \((-1, -1)\), where the absolute minimum occurs, and \((8, 2)\), where the absolute maximum occurs. Ensure these points are labeled on the graph.

Key Concepts

Critical Points

In calculus, **critical points** are values of a function where its derivative is zero or undefined. These points are vital in studying functions because they can help determine where a function reaches its extreme values, like maximum or minimum values.

For finding critical points, follow these simple steps:

• First, take the derivative of the function. The derivative helps us understand how the function's output changes with respect to changes in the input.
• Next, equate the derivative to zero and solve for the variable. This gives potential critical points where the slope of the tangent to the function is flat.
• Also, check where the derivative is undefined, as these points need to be considered as potential critical points too.

In our example, the function is given by \(h(x) = \sqrt[3]{x} = x^{1/3}\). The derivative calculated was \(h'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\). Here, \(h'(x)\) is undefined at \(x=0\), indicating a critical point at \(x=0\).

To find absolute extremum on a given interval, always look at these critical points along with the endpoints of the interval.

Absolute Extrema

**Absolute extrema** refers to the absolute maximum and minimum values a function takes on over a specified interval. These extremal values provide crucial information about the full range of a function.

To determine absolute extrema, consider the following steps:

• Firstly, evaluate the function at all critical points within the interval.
• Next, evaluate the function at the interval's endpoints.
• Finally, compare these evaluated values to identify the highest and lowest function values.

In our exercise, we evaluated \(h(x)\) at the critical point and the endpoints of the interval \([-1, 8]\):

• \(h(-1) = -1\)
• \(h(0) = 0\)
• \(h(8) = 2\)

After calculating these, we found that the absolute maximum value is \(2\), occurring at \(x = 8\), and the absolute minimum value is \(-1\) at \(x = -1\). These points give a clear picture of where the function reaches its peak and lowest values within that interval.

Remember, the absolute extrema may occur either at critical points or at the boundary of the interval.

Derivative Calculation

The process of **derivative calculation** is foundational in calculus, helping to understand the rate of change of a function concerning its input. Calculating derivatives allows us to uncover critical points and extremal behavior of functions.

For functions involving powers of \(x\), apply the power rule of differentiation, which states: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).

Applying this to our example, where \(h(x) = x^{1/3}\), the derivative is calculated as:

• \(h'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\)

The derivative expresses the function's rate of change and opens insights into where the function increases or decreases. Wherever the derivative equals zero or becomes undefined, these points are potential sites for critical points.

In summary, derivative calculation is essential for analyzing functions, assisting in identifying both critical points and potential extremal values. By following these steps, you will gain a deeper understanding of how a function behaves over its domain.