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Let the function f : [0, ¥) ® R be such that
For x > 0 and f (0) = 1. Then f (1) lies in the interval
f'(x)=(8/x^2+3x+4)?
For x > 0 and f (0) = 1. Then f (1) lies in the interval
f'(x)=(8/x^2+3x+4)?
Most Upvoted Answer
Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. ...
Explanation:
Given Information:
- Function f : [0, ¥) ® R
- f(x) = 1 for x > 0
- f(0) = 1
Analysis:
- The function f is defined for x > 0 and f(0) = 1.
- We are given the derivative of the function f, which is f'(x) = 8/x^2 + 3x + 4.
Applying Mean Value Theorem:
- By the Mean Value Theorem, there exists a c in (0,1) such that f'(c) = (f(1) - f(0))/(1-0).
- Since f(0) = 1, we have f'(c) = f(1) - 1.
Substitute f'(x) into the equation:
- Substitute the derivative f'(x) = 8/x^2 + 3x + 4 into the equation f'(c) = f(1) - 1.
- We get 8/c^2 + 3c + 4 = f(1) - 1.
Finding the interval for f(1):
- To find the interval for f(1), we need to determine the possible values of f(1) based on the given information.
- Since c lies in (0,1), we can analyze the values of the derivative f'(x) = 8/x^2 + 3x + 4 in this interval to determine the possible values of f(1).
Conclusion:
- By analyzing the derivative and applying the Mean Value Theorem, we can determine the interval in which f(1) lies based on the given information and the properties of the function f.
Given Information:
- Function f : [0, ¥) ® R
- f(x) = 1 for x > 0
- f(0) = 1
Analysis:
- The function f is defined for x > 0 and f(0) = 1.
- We are given the derivative of the function f, which is f'(x) = 8/x^2 + 3x + 4.
Applying Mean Value Theorem:
- By the Mean Value Theorem, there exists a c in (0,1) such that f'(c) = (f(1) - f(0))/(1-0).
- Since f(0) = 1, we have f'(c) = f(1) - 1.
Substitute f'(x) into the equation:
- Substitute the derivative f'(x) = 8/x^2 + 3x + 4 into the equation f'(c) = f(1) - 1.
- We get 8/c^2 + 3c + 4 = f(1) - 1.
Finding the interval for f(1):
- To find the interval for f(1), we need to determine the possible values of f(1) based on the given information.
- Since c lies in (0,1), we can analyze the values of the derivative f'(x) = 8/x^2 + 3x + 4 in this interval to determine the possible values of f(1).
Conclusion:
- By analyzing the derivative and applying the Mean Value Theorem, we can determine the interval in which f(1) lies based on the given information and the properties of the function f.
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Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)?
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Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)?.
Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the function f : [0, ¥) ® R be such that For x > 0 and f (0) = 1. Then f (1) lies in the interval f'(x)=(8/x^2+3x+4)?.
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