Rolle’s Theorem states that “Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b) then there is at least one number c in (a,b) such that f’(c) = 0.
Ex. Find the two x-intercepts of f(x) = x^2+5x+4
Set the equation equal to zero
F(x) = x^2+5x+4 = 0
Then factor the equation
F(x) = (x+4)(x+1) = 0
So, f(4) = f(1) = 0, and from Rolle’s Theorem you know that there exists at least one c in the interval (-1,-4) such that f’(c) = 0.
Take the derivative of the original equation
F(x) = x^2+5x+4 = 0
F’(x) = (2)x^2-1+5(1)x^1-1+4^0 = 0
F’(x) = 2x^1+5 = 0
And determine that f’(x) = 0 when x = -5/2, Note that this x-value lies in the open interval (-1,-5).
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