0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '16 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 '16 at 13:30
0
$\begingroup$

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: $0 < \frac{\cos(x) + x - 1}{x-0}$.

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ you need $1-\sin(c)>0$ (which follows from $0<c<\pi$). $\endgroup$ – LinAlg Sep 29 '16 at 13:56

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.