Chapter 4: Problem 21
Identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\sqrt[3]{x} ; I=[-1,27] $$
Short Answer
Expert verified
Minimum value: -1 at \( x = -1 \); Maximum value: 3 at \( x = 27 \).
Step by step solution
01
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, \( g'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \).
02
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 \( x = 0 \).
03
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 \( x = 27 \): \[ g(27) = \, \sqrt[3]{27} = 3 \]
04
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 \( x = 27 \): \( g(27) = 3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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: \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \).
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.
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 \( x = 27 \): \( g(27) = \sqrt[3]{27} = 3 \)
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.
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.