`f(x) = (x^2 - 2x - 3)/(x + 2), [-1,3]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c`...

`f(x) = (x^2 - 2x - 3)/(x + 2), [-1,3]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open interval

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Chapter 3, 3.2 - Problem 15 - Calculus of a Single Variable (10th Edition, Ron Larson).
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The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if f(-1) = f(3).

`f(-1) = ((-1)^2-2*(-1)-3)/(-1+2) = (1+2-3)/1 = 0`

`f(3) = ((3)^2-2*(3)-3)/(3+2) = (9-6-3)/5 = 0`

Since all the three conditions are valid, you may apply Rolle's theorem:

` f'(c)(b-a) = 0`

Replacing 3 for b and -1 for a, yields:

`f'(c)(3+1) = 0`

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

`f'(c) = ((c^2-2*c-3)'*(c+2) - (c^2-2*c-3)*(c+2)')/((c+2)^2) => f'(c) = ((2c-2)(c+2) -c^2 + 2c + 3)/((c+2)^2) `

` f'(c) = (2c^2 + 4c - 2c - 4 - c^2 + 2c + 3)/((c+2)^2) `

` f'(c) = (c^2 + 4c - 1)/((c+2)^2) `

Replacing the found values in equation 4f'(c) = 0

`4(c^2 + 4c - 1)/((c+2)^2)  = 0 => c^2 + 4c - 1 = 0`

`c_(1,2) = (-4+-sqrt(16+4))/2 => c_(1,2) = (-4+-2sqrt5)/2 =>  c_(1,2) = (-2+-sqrt5)`

Since `c = (-2-sqrt5) ` does not belong to (-1,3), only` c =(-2+sqrt5)` is a valid value.

Hence, in this case, the Rolle's theorem may be applied for `c = (-2+sqrt5).`

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