Determine whether the Mean Value theorem can be applied to f on the closed interval [a, b]. (Select all that a?  f(x) = square root (20 − x), [−5, 20]If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = f (b) − f (a)b − a. (Enter your answers as a comma-separated list. If the Mean Value Theorem cannot be applied, enter NA.)c =

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You need to use the mean value theorem for the function `f(x) = sqrt(20-x)` , over the interval `[-5,20` ] such that:

`f(20) - f(-5) = f'(c)(20 - (-5))`

You need to remember that `c in (-5,20)`  and you need to evaluate `f'(x), ` then you need to substitute c...

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You need to use the mean value theorem for the function `f(x) = sqrt(20-x)` , over the interval `[-5,20` ] such that:

`f(20) - f(-5) = f'(c)(20 - (-5))`

You need to remember that `c in (-5,20)`  and you need to evaluate `f'(x), ` then you need to substitute c for x in equation representing `f'(x) `  such that:

`f'(x) = (sqrt(20-x))' => f'(x) = ((20-x)')/(2sqrt(20-x))`

`f'(x) = -1/(2sqrt(20-x)) => f'(c) = -1/(2sqrt(20-c))`

You need to evaluate `f(20) ` and `f(-5) ` such that:

`f(20) = sqrt(20-20) = 0`

`f(-5) = sqrt(20+5) = sqrt25 = 5`

Substituting these values in equation representing the mean value theorem yields:

`0 - 5 = -25/(2sqrt(20-c)) => -5 = -25/(2sqrt(20-c))`

Reducing by -`5`  yields:

`1 = 5/(2sqrt(20-c)) => 2sqrt(20-c) = 5`

You need to raise to square to remove the square root such that:

`4(20-c) = 25 => 20 - c = 25/4 => c = 20 - 25/4 => c = 55/4 = 13.75in (-5,20).`

Hence, evaluating  `c in (-5,20),`  for the given function, using mean value theorem yields `c = 13.75` .

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