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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

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  • $\begingroup$ It's odd that the function is expressed as $x^2/x$ instead of just $x$. Are you sure you typed this correctly? $\endgroup$
    – B. Goddard
    Sep 29, 2016 at 13:28
  • $\begingroup$ I'm almost sure it should be $\frac{x^2}{2}$ and not $\frac{x^2}{x}$. $\endgroup$
    – Daniel Fischer
    Sep 29, 2016 at 13:30

2 Answers 2

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Note that it suffices to prove the inequality for on $[0,2]$. Apply the MVT with $a=0$ and $b=x$ ($x\leq2$) to obtain $-\sin(c)+1 = \frac{\cos(x) + x - 1}{x-0}$ for some $c \in (0,x)$. To get rid of $-\sin(c)+1$, we can either use $-\sin(c)+1<1$ or $-\sin(c)+1 > 0$. Since we want to end up with $\cos(x) > ...$, we use the latter: $0 < \frac{\cos(x) + x - 1}{x-0}$.

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You should check the application of the mean value theorem again, it is $\frac{f(x) - f(0)}{x}= 1 - \sin(c)$. You can then conclude by arguing that $1 - \sin(c) \ge 0$ for $c \le x \le \pi$. (Thanks for the pointer LinAlg)

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    $\begingroup$ you need $1-\sin(c)>0$ (which follows from $0<c<\pi$). $\endgroup$
    – LinAlg
    Sep 29, 2016 at 13:56

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