Question

Prove that |cos(x) - 1| ≤ |x| for all x. (Use Mean Value Theorem)

Answer 1

I'll just outline the idea of the proof:
first, there are definitely some values of x for which the inequality holds (e.g. pi/2)
now suppose there is a point x* such that the inequality is not true:
that implies that the LHS of the inequality goes from below the y=x line to above the y=x line. (using the language of mean value theorem, this means the secant line drawn from x=pi/2 to x = x* has a slope greater than 1). Now mean value theorem would say that there exists a point c such that f'(c) = slope of secant line > 1.
look at your function f(x) = cos(x) - 1, its derivative is ~sin(x), which is bound by -1 and 1.
and there's the contradiction that completes the proof.

Answer 2

Thanks for the detailed response cnknd!