Math Problem Statement
solve 29 and 31
Solution
To solve these problems, we need to use the fact that the derivative of the inverse function at a given point is:
We'll proceed with each question individually.
Problem 29
Given: and .
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Verify that has an inverse:
Since is a strictly increasing function (its derivative is positive for all ), is one-to-one and hence has an inverse. -
Find : We need to solve for such that : So, .
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Find :
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Evaluate : Substitute into : [ f'(f^{-1}(-1)) = f'(-\sqrt[3]{3}) = 3(-\sqrt[3]{3})^2 = 3 \cdot \sqrt[3]{9} ]
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Calculate :
So, the answer for problem 29 is:
Problem 31
Given: and .
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Verify that has an inverse:
Within the interval , is strictly increasing and continuous, thus it is one-to-one and has an inverse. -
Find : We need to solve for such that : The solution within the given interval is: Thus, .
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Find :
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Evaluate : Substitute into : Since ,
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Calculate :
So, the answer for problem 31 is:
Would you like a further breakdown of these steps, or do you have any other questions?
Related Questions
- How can we generalize the formula for the derivative of an inverse function?
- Why is the domain restriction necessary for to have an inverse?
- Can you explain why a function being strictly increasing implies it has an inverse?
- How would the derivative of the inverse change if the function were decreasing instead of increasing?
- What would be the steps to find the inverse derivative of a composite function?
Tip
When verifying if a function has an inverse, checking if it is strictly monotonic (either increasing or decreasing) over the entire domain can simplify the process.
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Math Problem Analysis
Mathematical Concepts
Inverse Functions
Derivatives
Trigonometry
Formulas
Derivative of an inverse function: (f^{-1})'(a) = 1 / f'(f^{-1}(a))
Derivative of cubic functions
Derivative of the tangent function: f'(x) = sec^2(x)
Theorems
Theorem on the derivative of inverse functions (Theorem 5.9)
Suitable Grade Level
Grades 11-12
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