Math Problem Statement
Solution
We are tasked with finding the absolute maximum and minimum of the function:
on the interval . Let's break this down step by step.
1. Domain of the Function
The given function involves a fractional exponent, so we need to ensure that the expression is well-defined. Since the exponent is positive, this expression is defined for all . However, for , the function is still real because fractional powers with even numerators (like ) still produce real results.
2. Critical Points (First Derivative Test)
To find critical points, we'll compute the derivative and set it equal to zero. However, based on the form of the function, it's important to focus on how the function behaves near and at the endpoints of the interval .
Evaluate at Key Points
We’ll evaluate at the endpoints of the interval and check for any critical points within the interval.
Next, we determine the behavior at , which is where the base of the power becomes zero:
3. Compare Values
After evaluating the function at critical points and endpoints, compare the values to find the absolute maximum and minimum.
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivative
Maxima and Minima
Critical Points
Formulas
f(x) = (x - 3)^(4/5) - 3
First Derivative Test: f'(x) = 0 to find critical points
Evaluating function at endpoints and critical points
Theorems
First Derivative Test for Critical Points
Absolute Maxima and Minima on a Closed Interval
Suitable Grade Level
Grades 11-12 (Calculus)
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