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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?

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  • $\begingroup$ The sign of $f'(c)$ gives the result (and the approach works for $x\lt0$ as well). $\endgroup$ – Did Aug 5 '14 at 9:47
  • $\begingroup$ Meaning that f′(c) is positive? $\endgroup$ – ALEXANDER Aug 5 '14 at 9:50
  • $\begingroup$ Tell me. $ $ $ $ $\endgroup$ – Did Aug 5 '14 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 __ $\endgroup$ – ALEXANDER Aug 5 '14 at 9:57
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    $\begingroup$ see it as $f(x)>[f(0)=1]$ $\endgroup$ – RE60K Aug 5 '14 at 11:17
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$ 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 $f(x)>f(0)=1$

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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

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

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  • $\begingroup$ Why not factor out the constant in the derivative $\endgroup$ – ALEXANDER Aug 5 '14 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 '14 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 '14 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 $\endgroup$ – ALEXANDER Aug 5 '14 at 10:19
  • $\begingroup$ That I can see, but I do not get where the number 1 is coming from. $\endgroup$ – ALEXANDER Aug 5 '14 at 10:22

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