Using mean value theorem to show that $\cos (x)>1-x^2/2$ I have a question, by applying the mean value theorem  to $f(x)=\frac{x^2}{2}+\cos (x)$, on the interval $[0,x]$, show that $\cos (x)>1-\frac{x^2}{2}$.
We know that $\frac{\text{df}(x)}{\text{dx}}=x-\sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.
This is where I get confused:
so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)\cdot x=1$ or what is going on? 

You started off well.
Notice that, by MVT:
$$f'(c) = \frac{f(x) - f(0)}{x - 0}$$
S0
$$xf'(c) = f(x) - f(0)$$
Notice that x is positive, and since $$f'(x) = x - sin(x)$$
Also, note that $x > \sin(x)$, so $f'(x) > 0$
Therefore,
We can conclude that
$$f(x) > f(0)$$
And
$$\cos(x) > 1- \frac{x^2}{2}$$
