You need to evaluate the derivative of the function f(x) = 2x - 2 - cos x , such that:
f'(x) = 2 + sin x
You need to use Rolle's theorem, so you need to find the roots of the equation 2 + sin x = 0.
sin x = -2
Notice that there is no value of x for sin x = -2 sin sin x in [-1,1], hence, the equation sin x + 2 = 0 has no solution. Since the equation f'(x) = 0 has no solution, there is no change of sign for the function f(x), over the interval (-oo,+oo), hence, the equation f(x) = 0 has no real solutions.
Thursday, July 26, 2012
Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 68
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