Determine whether the Mean Value Theorem applies to the function f(x)= x+1/x on the interval [2,6]. If so find or approximate the point(s) that are guaranteed by the existing Mean Value Theorem.

Answer 1

I'll assume you mean f(x to be:

There are two criteria for the Mean Value Theorem to be valid. 1) the function must me continuous on the interval and 2) the function must be differentiable on the interval.

So for criterion 1) Is the function continuous on [2, 6]. Yes. This function is not continuous at x = 0 but that's not on the interval [2, 6] so, no problem.

For criterion 2) is the function differentiable on the interval? Well f '(x) requires us to use the quotient rule (u/v)' = (u'v - uv')/v² so f '(x) = (x - (x+1))/x² = -1/x² and this is valid for all "x" except x = 0 so, again, no problem.

According to the mean value theorem, there must be some value between x = 2 an x = 6 where the slope of the tangent line to the curve is equal to the (f(6) - f(2))/(6 - 2)

f(6) = 7/6 and f(2) = 3/2 so f(6) - f(2) = 7/6 - 3/2 = 7/6 - 9/6 = -2/6 = -1/3

So (f(6) - f(2))/(6 - 2) = (-1/3)/(4) = -1/12

Since the derivative is -1/x² we can set that equal to - 1/12 to find the associated value of x:

-1/12 = -1/x²

x² = 12

x = √12 (approximately 3.464)

So at x = √12 the slope of the tangent line equals the slope of the secant line between (2, f(2)) and (6, f(6))