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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