Math Problem Statement
Solution
To solve for the absolute maximum and absolute minimum of the function on the interval , we will follow these steps:
Step 1: Find the derivative of the function
First, we need to find the critical points by taking the derivative of .
The derivative is:
Simplifying:
Step 2: Set the derivative equal to 0 to find critical points
We solve :
Cubing both sides:
So, the critical points are and .
Step 3: Evaluate the function at critical points and endpoints
We now evaluate at the critical points and the endpoints and .
Step 4: Determine absolute minimum and maximum
- The minimum value is .
- The maximum value is .
Thus:
- The absolute minimum is approximately .
- The absolute maximum is approximately .
Let me know if you need further clarification or additional details!
Follow-up questions:
- What are the exact critical points of a function with a cubic root?
- Why are the endpoints also evaluated for absolute extrema?
- How does the cubic root affect the function's behavior near zero?
- Can the second derivative test confirm the nature of the critical points here?
- How would you apply the same method to a different interval, e.g., ?
Tip:
Always check both the critical points and the boundary points of the interval when looking for absolute extrema!
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Math Problem Analysis
Mathematical Concepts
Calculus
Critical Points
Derivatives
Extrema
Formulas
f(t) = t - ³√t
f'(t) = 1 - 1/(3√t²)
Theorems
First Derivative Test
Critical Point Theorem
Suitable Grade Level
College level (Calculus I)
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