Some calculus quesitons

a)
Let f(x) = x^2 - cos x -1
f'(x) = 2x + sin x
The function f'(x) is differentiable on R.
Moreover, its derivative f"(x) = 2 + cos x > 0, for all x (- R.
Therefore, f'(x) is increasing.
Since f'(0) = 0, one gets f '(x) > 0 for all x > 0

b)

f(1) = -0.999847695
f(3/2) = 0.250342675

So there is at least one root in (1, 1.5).
As f'(x) is strictly increasing in (1, 1.5) thus there can be only one root in (1, 1.5)

2012-10-30 01:15:28 補充：
Ref
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-2249.png