Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\sqrt[3]{x} ; I=[-1,27] $$

Step-by-Step Solution

Answer

Minimum value: -1 at \( x = -1 \); Maximum value: 3 at \( x = 27 \).

Step 1: Find the Derivative

First, we need to find the derivative of the function to determine the critical points. The function is given as \( g(x) = x^{1/3} \). Using the power rule for derivatives, \( g'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \).

Step 2: Determine Critical Points

Critical points occur where the derivative is zero or undefined. The derivative \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \) is undefined when \( x = 0 \), because the denominator becomes zero. There are no points where the derivative is zero because the numerator is a constant (1). Thus, the critical point is at \( x = 0 \).

Step 3: Evaluate the Endpoints and Critical Points

Now, evaluate the function at the endpoints and the critical point found in the interval \([-1, 27]\). At \( x = -1 \): \[ g(-1) = \, \sqrt[3]{-1} = -1 \]At the critical point \( x = 0 \):\[ g(0) = \, \sqrt[3]{0} = 0 \]At \( x = 27 \): \[ g(27) = \, \sqrt[3]{27} = 3 \]

Step 4: Identify Maximum and Minimum Values

To find the maximum and minimum values, compare \( g(x) \) at the evaluated points.- Minimum value is at \( x = -1 \): \( g(-1) = -1 \).- Maximum value is at \( x = 27 \): \( g(27) = 3 \).

Key Concepts

Derivative Calculation

Understanding how to calculate derivatives is a fundamental skill in calculus. For the function \( g(x) = x^{1/3} \), we utilize the power rule to find its derivative. The power rule states that if you have a function \( f(x) = x^n \), the derivative \( f'(x) \) is \( n \cdot x^{n-1} \).

For our function, the exponent \( 1/3 \) is applied:

• Using the power rule: \( g'(x) = \frac{1}{3}x^{-2/3} \).
• This expression can be rewritten as: \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \).

Derivatives help us find rates of change, slopes of tangent lines, and are essential in identifying critical points where certain behaviors or properties of a function change.

Maximum and Minimum Values

To discover where a function reaches its highest or lowest points within a given interval, we look for critical points and evaluate the endpoints. A function can achieve these values at:

• Critical points: where the derivative is zero or undefined.
• Endpoints: the edges of a given interval.

The function \( g(x) = x^{1/3} \) has its derivative undefined at \( x = 0 \), marking it as a critical point. We then evaluate \( g(x) \) at the critical point and the endpoints to compare these values. This helps to determine the absolute maximum and minimum values within the interval.

Function Evaluation

Function evaluation involves substituting specific values into a function to determine its output at those points. This step is crucial for comparing potential maximum and minimum values. We evaluate \( g(x) \) at crucial points:

• At endpoint \( x = -1 \): \( g(-1) = \sqrt[3]{-1} = -1 \)
• At critical point \( x = 0 \): \( g(0) = \sqrt[3]{0} = 0 \)
• At endpoint \( x = 27 \): \( g(27) = \sqrt[3]{27} = 3 \)

These evaluations allow us to see which points correspond to the highest or lowest values of the function over the interval.

Endpoints Analysis

Evaluating endpoints is essential in determining the extreme values of a function on a closed interval. They represent the boundaries within which a function's behavior is contained.
The interval \([-1, 27]\) means you evaluate \( g(x) \) at \( x = -1 \) and \( x = 27 \) in addition to any critical points.

• At \( x = -1 \): \( g(-1) = -1 \), representing one of the lower boundary values.
• At \( x = 27 \): \( g(27) = 3 \), indicating the upper boundary value.

By analyzing these endpoints in conjunction with critical points, you can determine the absolute maximum and minimum values of the function over the specified interval. This ensures that all potential extreme values are considered.