`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
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.