`f(x) = sqrt(2 - x), [-7,2]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in...

`f(x) = sqrt(2 - x), [-7,2]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. 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)`. If the Mean Value Theorem cannot be applied, explain why not.

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Textbook Question

Chapter 3, 3.2 - Problem 44 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.

The mean value theorem states:

`f(b) - f(a) = f'(c)(b-a)`

Replacing 2 for b and -7 for a, yields:

`f(2) - f(-7) = f'(c)(2+7)`

Evaluating f(2) and f(-7) yields:

`f(2) =sqrt(2-2) = 0`

`f(-7) = sqrt(2+7)=3`

You need to evaluate f'(c), using quotient rule:

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

Replacing the found values in equation `f(2) - f(-7) = f'(c)(2+7):`

`0 - 3= -9/(2sqrt(2-c)) =>3/(2sqrt(2-c)) = 1 =>2sqrt(2-c) = 3=> sqrt(2-c) = 3/2 => 2 - c = 9/4 => c = 2 - 9/4 => c = -1/4 in (-7,2)`

Hence, in this case, the mean value theorem may be applied for `c = -1/4.`

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