If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:2:54

AP.CALC:

FUN‑1 (EU)

, FUN‑1.B (LO)

, FUN‑1.B.1 (EK)

- [Tutor] Let g of x equal one over x. Can we use the mean value theorem to say that the equation g
prime of x is equal to one half has a solution where negative one is less than x is less than two, if so, write a justification. Alright, pause this video and see if you can figure that out. So the key to using
the mean value theorem, even before you even think about using it, you have to make sure that you are continuous
over the closed interval and differentiable over the open interval, so this is the open interval here and then the closed interval
would include the end points. But you might immediately realize that both of these intervals
contain x equals zero and if x equals zero,
the function is undefined and if it's undefined there, well it's not going to continuous or differentiable at that point and so no, not continuous or differentiable, differentiable over the interval, over the interval. Alright, let's do the second part. Can we use the mean value theorem to say that there is a value
c such that g prime of c is equal to negative one half and one is less than c is less than two if so, write a justification. So pause the video again. Alright, so in this
situation, between one and two on both the open and the closed intervals, well, this is a rational function and a rational function
is going to be continuous and differentiable at
every point in its domain and its domain completely contains this open and closed interval or another way to think about it, every point on this open interval and on the closed
interval is in the domain, so we can write g of x is a rational function, which lets us know that it is continuous and differentiable at
every point in its domain, at every point in its domain, the closed interval from one to two is in domain and so now, let's see what
the average rate of change is from one to two and so we get g of two minus g of one over two minus one is equal to one half minus one over one, which is
equal to negative one half, therefore, therefore by the mean value theorem, there must be a c where one is less than c is less than two and g prime of c is equal to the average rate of change
between the end points, negative one half and we're done, so we can put
a big yes right over there and then this is our justification.