# 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).