# f(x)=ln(4+2x-x^2); [-1,3] using rolle's theorem

f(x)=ln(4+2x-x^2); [-1,3] using rolle's theorem

We can only log positive values, so the polynomial 4 + 2x - x^2 within the logarithmn above must be more than zero. We have:

4 + 2x - x^2 > 0
x^2 - 2x - 4 < 0
(x - 1)^2 - 5 < 0
(x - 1)^2 < 5
-sqrt(5) < x - 1 < sqrt(5)
1 - sqrt(5) < x < 1 + sqrt(5)

Thus, the domain of f(x) is (1 - sqrt(5) , 1 + sqrt(5)).
1 - sqrt(5) < -1 < 3 < 1 + sqrt(5), so [-1, 3] is a subinterval of the domain of f(x).

logarithmic functions are continuous and differentiable on their domain, and since [-1, 3] is a subinterval of f(x)'s domain, f(x) must be continuous on [-1, 3] and differentiable on (-1, 3).

f(-1) = ln (4 + 2(-1) - (-1)^2) = ln (1) = 0
f( 3) = ln (4 + 2(3) - 3^2) = ln (1) = 0

Hence, by rolle's theorem, there must exists c in the interval (-1 , 3) such that f'(c) = 0.
answered Apr 25, 2012 by anonymous

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