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The mean value c for the function f(x) = logx and g(x) = log(1/X) in [1, 2] is
- a)1 only
- b)1.5 only
- c)Any value between 1 and 2
- d)1.25
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The mean value c for the functionf(x) = logx and g(x) = log(1/X) in [1...
Both the functions are continuous in [1, 2] and differentiable in (1, 2).
By Cauchy's Mean Value theorem we should have
Therefore c can be any value between 1 and 2.
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Community Answer
The mean value c for the functionf(x) = logx and g(x) = log(1/X) in [1...
Mean Value Theorem:
The mean value theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].
Given Functions:
We are given two functions: f(x) = log(x) and g(x) = log(1/x).
Step 1: Check if the Functions are Continuous and Differentiable:
To apply the mean value theorem, we need to check if both functions are continuous and differentiable on the interval [1, 2].
1. f(x) = log(x):
- The function f(x) = log(x) is continuous on the interval [1, 2] since the logarithmic function is continuous for positive values of x.
- The function f(x) = log(x) is differentiable on the interval (1, 2) since the derivative of the logarithmic function exists for positive values of x.
2. g(x) = log(1/x):
- The function g(x) = log(1/x) is continuous on the interval [1, 2] since the logarithmic function is continuous for positive values of x.
- The function g(x) = log(1/x) is differentiable on the interval (1, 2) since the derivative of the logarithmic function exists for positive values of x.
Therefore, both functions f(x) = log(x) and g(x) = log(1/x) satisfy the conditions of the mean value theorem on the interval [1, 2].
Step 2: Find the Derivatives of the Functions:
To find the derivative of f(x) = log(x), we can use the logarithmic differentiation:
f'(x) = (1/x)
To find the derivative of g(x) = log(1/x), we can use the chain rule:
g'(x) = (1/(1/x)) * (-1/x^2) = -1
Step 3: Apply the Mean Value Theorem:
Since both functions f(x) = log(x) and g(x) = log(1/x) satisfy the conditions of the mean value theorem on the interval [1, 2], there exists at least one value c in the interval (1, 2) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [1, 2].
The average rate of change of f(x) = log(x) over the interval [1, 2] is:
(f(2) - f(1))/(2 - 1) = (log(2) - log(1))/(2 - 1) = log(2)
The average rate of change of g(x) = log(1/x) over the interval [1, 2] is:
(g(2) - g(1))/(2 - 1) = (log(1/2) - log(1))/(2 - 1) = -log(2)
Since the derivative of f(x) = log(x) is 1/x and the derivative of g(x) = log(
The mean value theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].
Given Functions:
We are given two functions: f(x) = log(x) and g(x) = log(1/x).
Step 1: Check if the Functions are Continuous and Differentiable:
To apply the mean value theorem, we need to check if both functions are continuous and differentiable on the interval [1, 2].
1. f(x) = log(x):
- The function f(x) = log(x) is continuous on the interval [1, 2] since the logarithmic function is continuous for positive values of x.
- The function f(x) = log(x) is differentiable on the interval (1, 2) since the derivative of the logarithmic function exists for positive values of x.
2. g(x) = log(1/x):
- The function g(x) = log(1/x) is continuous on the interval [1, 2] since the logarithmic function is continuous for positive values of x.
- The function g(x) = log(1/x) is differentiable on the interval (1, 2) since the derivative of the logarithmic function exists for positive values of x.
Therefore, both functions f(x) = log(x) and g(x) = log(1/x) satisfy the conditions of the mean value theorem on the interval [1, 2].
Step 2: Find the Derivatives of the Functions:
To find the derivative of f(x) = log(x), we can use the logarithmic differentiation:
f'(x) = (1/x)
To find the derivative of g(x) = log(1/x), we can use the chain rule:
g'(x) = (1/(1/x)) * (-1/x^2) = -1
Step 3: Apply the Mean Value Theorem:
Since both functions f(x) = log(x) and g(x) = log(1/x) satisfy the conditions of the mean value theorem on the interval [1, 2], there exists at least one value c in the interval (1, 2) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [1, 2].
The average rate of change of f(x) = log(x) over the interval [1, 2] is:
(f(2) - f(1))/(2 - 1) = (log(2) - log(1))/(2 - 1) = log(2)
The average rate of change of g(x) = log(1/x) over the interval [1, 2] is:
(g(2) - g(1))/(2 - 1) = (log(1/2) - log(1))/(2 - 1) = -log(2)
Since the derivative of f(x) = log(x) is 1/x and the derivative of g(x) = log(
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The mean value c for the functionf(x) = logx and g(x) = log(1/X) in [1, 2] isa)1 onlyb)1.5 onlyc)Any value between 1 and 2d)1.25Correct answer is option 'C'. Can you explain this answer?
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