Math Problem Statement
Solution
The problem asks for the value of where the function has a critical point. To find the critical point, we need to compute the derivative , set it equal to zero, and solve for .
Steps:
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Compute the first derivative : The function is , where . Using the chain rule: Now, find :
Therefore, the derivative becomes:
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Set : Since for any real , the critical points occur when:
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Solve for :
Thus, the function has a critical point when .
Would you like more details or have any further questions? Here are 5 related questions to consider:
- What is the definition of a critical point in calculus?
- How does the chain rule apply when differentiating exponential functions?
- Can be any value, and how does it affect the critical point?
- What does the behavior of the second derivative tell us about the nature of this critical point?
- How would this problem change if involved different exponents or terms?
Tip: To analyze critical points thoroughly, always check both the first and second derivatives for sign changes and concavity.
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Math Problem Analysis
Mathematical Concepts
Calculus
Differentiation
Critical Points
Formulas
f'(x) = e^g(x) * g'(x)
g'(x) = x^9 + 11p
f'(x) = 0 => x^9 + 11p = 0
Theorems
Chain Rule
Suitable Grade Level
College level (Calculus 1 or 2)