Math Problem Statement
The function f(x)=8x+2x^-1 has one local minimum and one local maximum. This function has a local maximum at x= with value
and a local minimum at x= with value
Solution
To find the local maximum and minimum of the function , let's go through the steps:
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Find the derivative to locate critical points: Differentiating with respect to :
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Set to find critical points: Solving for :
So, the critical points are and .
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Determine if these points are maxima or minima using the second derivative :
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At : Since , is a local minimum.
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At : Since , is a local maximum.
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Calculate the function values at these points:
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For the local minimum at :
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For the local maximum at :
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Summary of Results:
- The local maximum occurs at with value .
- The local minimum occurs at with value .
Would you like a more detailed explanation on any of these steps?
- What is the significance of the second derivative test for identifying maxima and minima?
- How would the result change if we modified the function to ?
- What happens to as approaches 0?
- How do we interpret the local maxima and minima on a graph?
- Can this method be applied to find inflection points?
Tip: When differentiating functions with terms like , rewriting them as fractions or powers simplifies the derivative process.
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Math Problem Analysis
Mathematical Concepts
Calculus
Differentiation
Critical Points
Second Derivative Test
Formulas
f(x) = 8x + 2x^-1
f'(x) = 8 - 2x^-2
f''(x) = 4/x^3
Theorems
Second Derivative Test
Suitable Grade Level
Grades 11-12
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