`f(x) = (x + 1)/x, [-1,2]` 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) = (x + 1)/x, [-1,2]` 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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Chapter 3, 3.2 - Problem 42 - 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 1 for b and 0 for a, yields:

`f(2) - f(-1) = f'(c)(2+1)`

Evaluating f(2) and f(-1) yields:

`f(2) =(2+1)/2 => f(2) =3/2`

`f(-1) = (-1+1)/(-1) => f(-1) = 0`

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

`f'(c) = ((c+1)/c)' => f'(c) = ((c+1)'*c - (c+1)*c')/(c^2) => f'(c) = (c - c - 1)/(c^2) => f'(c) = -1/(c^2)`

Replacing the found values in equation `f(2) - f(-1) = f'(c)(2+1):`

`3/2-0= -3/(c^2) =>1/2 =-1/(c^2)=>c^2 = -2 !in R`

Hence, in this case, there is no valid c value for the mean value theorem to be applied.

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