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# Justification with the mean value theorem: table

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.B (LO)
,
FUN‑1.B.1 (EK)

## Video transcript

the table gives selected values of the differentiable function f alright can we use a mean value theorem to say that there is a value C such that F prime of C is equal to 5 and C is between 4 & 6 if so right of justification well it meets to meet the to use the mean value theorem you have to be differentiable over the open interval and continuous over the closed interval so it seems like we've met that because if you're differentiable over an interval you're definitely can continuous over that interval that's saying that's just a generally differentiable function f I guess over any interval but the next part is to say alright but if that condition is met then the slope of the secant line between 4 comma F of 4 and 6 comma f of 6 that sum at least one point in between 4 & 6 will have a derivative that is equal to the slope of the secant line and so let's figure out what the slope of the secant line is between 4 comma F of 4 and 6 comma f of 6 and if it's equal to 5 then we could use the mean value theorem if it's not equal to 5 then the mean value theorem would not apply and so let's do that F of 6 minus F of 4 all of that over 6 minus 4 is equal to 7 minus 3 over 2 which is equal to 2 so 2 not equal to 5 so mean value theorem doesn't apply alright let's put an exclamation mark there from mark there for emphasis I let's do the next part can we use the mean value theorem to say that the equation F prime of X is equal to negative 1 has a solution and now the interval is from 0 to 2 if so write a justification all right so let's see this so if we were to take the slope of the secant line so f of 2 minus f of 0 all that over 2 minus 0 this is equal to negative 2 minus 0 all of that over 2 which is equal to negative 2 over 2 which is equal to negative 1 and so and we also know that we meet the continuity and differentiability conditions and so we could say and since F is generally just French differentiable generally differentiable differentiable it will be differentiable differentiable and and continuous over the interval from zero to two and I say the close interval you just have to be differentiable over the open interval but it's even better I guess if you're differentiable over the closed interval cuz you have to be continuous over the closed interval and so and since F is generally differentiable it will be differentiable and continuous over zero too so the mean value theorem tells us tells us that there is an X in that interval from zero to two such that F prime of X is equal to that secant slope or you could say that average rate of change is equal to negative one and so I could write yes yes and then this would be my justification this is the slope of the secant line or the average rate of change and since F is generally differentiable it will be differentiable and continuous over the closed interval so the mean value theorem tells us that there is an X in this interval such that f prime of X is equal to negative one and we're done
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