Find the value of c that satisfies the Mean Value Theorem for f(x)=ln(x2) on the interval [1,e] . (1 point) Responses c=e−2 c = e − 2 c=e−1 c = e − 1 c=2e−2 c = 2 e − 2 c=2e−1 c = 2 e − 1 c=e+1
To apply the Mean Value Theorem, we need to verify that the function is continuous on the interval [1, e] and differentiable on the open interval (1, e).
First, let's find the derivative of f(x) = ln(x^2):
f'(x) = (1/x^2)(2x) = 2/x
Now we need to check the conditions of the Mean Value Theorem:
1. Continuity: The function f(x) = ln(x^2) is continuous on [1, e] since it is the composition of continuous functions (the natural logarithm and the square function), and these functions are continuous everywhere.
2. Differentiability: The function f(x) = ln(x^2) is differentiable on (1, e) since its derivative 2/x is defined and continuous on this interval.
Now let's find the average rate of change of f(x) = ln(x^2) on the interval [1, e]:
Average rate of change = [f(e) - f(1)] / (e - 1)
Average rate of change = [ln(e^2) - ln(1^2)] / (e - 1)
Average rate of change = [ln(e^2) - ln(1)] / (e - 1)
Average rate of change = [2ln(e) - ln(1)] / (e - 1)
Average rate of change = [2 - 0] / (e - 1)
Average rate of change = 2 / (e - 1)
To use the Mean Value Theorem, we need to find a value of c in the open interval (1, e) such that f'(c) = 2 / (e - 1).
Setting f'(c) = 2 / (e - 1):
2/c = 2 / (e - 1)
c = e - 1
Therefore, the value of c that satisfies the Mean Value Theorem for f(x) = ln(x^2) on the interval [1, e] is c = e - 1.