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?

  • $\begingroup$ The sign of $f'(c)$ gives the result (and the approach works for $x\lt0$ as well). $\endgroup$
    – Did
    Aug 5, 2014 at 9:47
  • $\begingroup$ Meaning that f′(c) is positive? $\endgroup$
    Aug 5, 2014 at 9:50
  • $\begingroup$ Tell me. $ $ $ $ $\endgroup$
    – Did
    Aug 5, 2014 at 9:53
  • $\begingroup$ @Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __ $\endgroup$
    Aug 5, 2014 at 9:57
  • 1
    $\begingroup$ see it as $f(x)>[f(0)=1]$ $\endgroup$
    – RE60K
    Aug 5, 2014 at 11:17

2 Answers 2


$ f(x)=x^2/2+\cos(x)$

Note that $f(0)=0^2/2+1=1$ From your equation: $$f(x)-f(0)=(x)f'(c)=x(c-\sin c)$$

Let $g(x)=x-\sin x$ Again you can show that $g'(x)=1-\cos x$ which is always greater than $0$ due to bounded nature of $\cos x$.As $g(0)=0$ and it is an increasing function $\{g'(x)>0\;\forall x>0$}, thus $g(x)>0 \;\forall x>0$.

So $f(x)-f(0)>0\;\forall x>0$ as $x>0$ and $c-\sin c >0\;\forall c>0${as $0<c<x$}.

So $f(x)>f(0)=1$


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 $f'(x) > 0$


We can conclude that

$$f(x) > f(0)$$


$$\cos(x) > 1- \frac{x^2}{2}$$

  • $\begingroup$ Why not factor out the constant in the derivative $\endgroup$
    Aug 5, 2014 at 10:16
  • $\begingroup$ @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out $\endgroup$
    – Varun Iyer
    Aug 5, 2014 at 10:17
  • $\begingroup$ @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$ $\endgroup$
    – Varun Iyer
    Aug 5, 2014 at 10:19
  • $\begingroup$ You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant $\endgroup$
    Aug 5, 2014 at 10:19
  • $\begingroup$ That I can see, but I do not get where the number 1 is coming from. $\endgroup$
    Aug 5, 2014 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.