I'm having difficulties on this question:
By applying the Mean Value Theorem to $f(x)=\cos(x)+ \frac{x^2}{x}$ on the interval $[0, x]$ show that $\cos(x) > 1 - \frac{x^2}{x}$.
So far I've used $f'(c) = \frac{f(b)-f(a)}{b -a}$ and got to $-x(\sin(c)-c)=x-1$ but I have no idea how to go any further.
I'd appreciate any help.
Thanks