f(x) = ln x, [1, 6]
If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).
(ln6 - ln1)/(6-1)
ln6/5
So, we want ln c = ln6/5
c = e^(ln6/5) = 61/5
To determine the numbers that satisfy the conclusion of the Mean Value Theorem, we need to check if the function is continuous on the closed interval [1, 6] and differentiable on the open interval (1, 6).
First, let's check the continuity of f(x) = ln x on the interval [1, 6].
The natural logarithm function is continuous for positive values of x, which means that f(x) = ln x is continuous on the interval [1, 6] since both endpoints, 1 and 6, are positive.
Next, let's check the differentiability of f(x) = ln x on the interval (1, 6).
The derivative of f(x) = ln x is given by f'(x) = 1/x. Since 1/x is defined for all x on the interval (1, 6), f(x) = ln x is differentiable on (1, 6).
Therefore, f(x) = ln x satisfies the hypotheses of the Mean Value Theorem on the interval [1, 6].
According to the Mean Value Theorem, there exists at least one number c in the interval (1, 6) such that f'(c) = (f(6) - f(1))/(6 - 1).
Let's find this number c:
f(6) = ln 6,
f(1) = ln 1 = 0.
f'(c) = (f(6) - f(1))/(6 - 1)
= (ln 6 - 0)/(6 - 1)
= ln 6/5.
Therefore, the number c that satisfies the conclusion of the Mean Value Theorem is c = ln 6/5.
To apply the Mean Value Theorem (MVT), we need to check if the function f(x) = ln(x) satisfies the hypotheses of the theorem on the interval [1, 6]. The hypotheses are as follows:
1. Continuity: The function needs to be continuous on the closed interval [1, 6].
2. Differentiability: The function needs to be differentiable on the open interval (1, 6).
First, we check if f(x) = ln(x) is continuous on [1, 6]. The natural logarithm function, ln(x), is continuous for all positive values of x. Since the interval [1, 6] lies entirely in the positive range, f(x) = ln(x) is continuous on [1, 6].
Next, we check if f(x) = ln(x) is differentiable on (1, 6). The derivative of ln(x) is given by f'(x) = 1/x. We can see that f'(x) is defined for all positive values of x, including the interval (1, 6). Therefore, f(x) = ln(x) is differentiable on (1, 6).
Since f(x) = ln(x) satisfies the hypotheses, we can apply the Mean Value Theorem. According to the MVT:
f'(c) = (f(b) - f(a))/(b - a)
where c is some number in the open interval (1, 6), a = 1, and b = 6.
Now, we can find the numbers c that satisfy the conclusion of the Mean Value Theorem by finding the value of f'(c):
f'(c) = (f(6) - f(1))/(6 - 1)
= (ln(6) - ln(1))/(6 - 1)
To simplify this expression, we can use log properties. ln(1) = 0, so the numerator simplifies to ln(6). Dividing ln(6) by 5, we get:
f'(c) = ln(6)/5
Therefore, any number c in the open interval (1, 6) that gives f'(c) = ln(6)/5 satisfies the conclusion of the Mean Value Theorem.
In summary, the numbers c that satisfy the conclusion of the Mean Value Theorem for the function f(x) = ln(x) on the interval [1, 6] are given by:
c = (ln(6))/5