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)

Video transcript

let G of X equal 1 over X can we use the mean value theorem to say that the equation G prime of X is equal to 1/2 has a solution where negative 1 is less than X is less than 2 if so write a justification all right pause this video and see if you can figure that out so the key to using the mean value theorem 2 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 endpoints but you might immediately realize that both of these intervals contain x equals 0 and at x equals 0 the function is undefined and if it's undefined there well it's not going to be continuous or differentiable at that point and so no not continuous or differentiable differentiable over the interval over the interval all right let's do the second part can we use a mean value theorem to say that there is a value C such that G prime of C is equal to negative 1/2 and 1 is less than C is less than 2 if so write a justification so pause the video again alright so in this situation between 1 & 2 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 1 to 2 is in domain and so now let's see what the average rate of change is from 1 to 2 and so we get G of 2 minus G of 1 over 2 minus 1 is equal to 1/2 minus 1 over 1 which is equal to negative 1/2 therefore therefore by the mean value theorem there must be a C where 1 is less than C is less than 2 and G prime of C is equal to the average rate of change between the endpoints negative 1/2 and we're done so we could put a big yes right over there and then this is our justification
AP® is a registered trademark of the College Board, which has not reviewed this resource.