The geometric mean of two postitive numbers a and b is sqrt(ab).
Show that for f(x) = 1/x on any interval [a,b] of positive numbers, the value of c in the conclusion of the mean value theorem is c = sqrt(ab)
I have no idea how to do this!
If the mean of a and b is sqrt(ab), then (a + b) / 2 = sqrt(ab).
Is c supposed to be the area under the graph in interval [a,b] ? If so, calculate the antiderivative of f(x) and you'll have the formula for c.
It is the geometric mean. Isn't (a+b)/2 the arithmetic mean?
etyhw
need the geometric mean for 4 and 6
f(b)-f(a)
________ = f'(c)
b-a
1/b - 1/a
__________ = -1/(c)^2
b-a
You are correct, (a+b)/2 is the arithmetic mean, not the geometric mean. I apologize for the confusion.
To show that for f(x) = 1/x on any interval [a,b] of positive numbers, the value of c in the conclusion of the mean value theorem is c = sqrt(ab), we need to verify two conditions:
1. f(x) is continuous on the interval [a, b].
2. f(x) is differentiable on the open interval (a, b).
Let's check these conditions:
1. Continuity: The function f(x) = 1/x is continuous on any interval (a, b) where a and b are positive numbers, except at x = 0. However, since we are concerned with the interval [a, b], and if a and b are both positive, then f(x) is continuous on this interval.
2. Differentiability: The function f(x) = 1/x is differentiable on any interval (a, b) where a and b are positive numbers, except at x = 0. Again, since we are concerned with the interval [a, b], and if a and b are both positive, then f(x) is differentiable on this interval.
Therefore, the given function f(x) = 1/x satisfies the conditions of the mean value theorem on the interval [a, b].
According to the mean value theorem, for any function f(x) that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one value c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In this case, f(x) = 1/x on the interval [a, b], so f'(x) = -1/x^2.
Using the mean value theorem equation and substituting the values, we get:
-1/c^2 = (1/b - 1/a) / (b - a)
To simplify, let's cross-multiply and rearrange the equation:
-1/c^2 = (a - b) / (ab(b - a))
Let's multiply both sides by -c^2 to eliminate the negative sign:
c^2 = (b - a) / (ab(b - a))
Canceling out the (b - a) terms, we have:
c^2 = 1 / ab
Taking the square root of both sides, we get:
c = sqrt(1/ab)
And simplifying further, we have:
c = sqrt(ab)^-1
Therefore, c = sqrt(ab), which is the geometric mean of a and b.
So, we have shown that for the function f(x) = 1/x on any interval [a, b] of positive numbers, the value of c in the conclusion of the mean value theorem is c = sqrt(ab).