`f(x) = (x - 4)(x + 2)^2, [-2,4]` 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...

`f(x) = (x - 4)(x + 2)^2, [-2,4]` 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 12 - 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(-2) = f(4).

`f(-2) =(-2-4)(-2+2)^2 = -6*0 = 0`

`f(4) = (4-4)(4+2)^2 = 0*6^2 = 0`

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

` f'(c)(b-a) = 0`

Replacing 4 for b and -2 for a, yields:

`f'(c)(4+2) = 0`

You need to evaluate f'(c), using product and chain rules:

`f'(c) = (c-4)'(c+2)^2 + (c-4)*((c+2)^2)' => f'(c) = (c+2)^2 + 2(c+2)(c-4)`

Factoring out c+2 yields:

`f'(c) = (c+2)(c+2+2c-8)`

`f'(c) = (c+2)(3c-6)`

Replacing the found values in equation 6f'(c) = 0

`6(c+2)(3c-6) = 0 => (c+2)(c-2) = 0 => c = -2 and c = 2 !in R`

Since c = -2 does not belong to (-2,4), only c = 2 is a valid value.

Hence, in this case, the Rolle's theorem may be applied for c = 2.

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