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Let $f(x) = \sin x − x \cos x$, $0 \le x \le \pi$. Find the absolute maximum and the absolute minimum of f. Hence, or otherwise, determine the range of f. Finally, determine whether f has an inverse or not. You need not find the formula of the inverse function if exists.

I can find that the critical points are at $x=0$ and $x=\pi$ but when I do the sign test the abs max and min are also =0 What does this mean?

mathcounterexamples.net
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Be Happy
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  • Are you sure the max and min are 0? Did you plug the values of the critical points in the function? Sounds like you plugged them in the derivative.

    By the way, are you sure $x=\pi$ is a critical point? It is still a value to be checked (since it is an endpoint), but $f'(\pi)\neq 0$.

     – bartgol Nov 30 '15 at 15:46
  • @bartgol $f'(x) = x \sin (x) \implies f'(\pi) = 0$.. – Matthew Cassell Nov 30 '15 at 15:48
  • Oh, I did not see $\cos x$... My bad. As of the fact that the max and min are 0, I believe the problem is that the values were plugged in $f'$ rather than $f$ (and of course the result was 0$... – bartgol Nov 30 '15 at 15:59

2 Answers2

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$f$ is continuous, $\lim\limits_{k \to +\infty} f(2k \pi)=-\infty$ and $\lim\limits_{k \to -\infty} f(2k \pi)=+\infty$. Hence the range of $f$ is $\mathbb R$.

Also, $f$ vanishes for an infinite number of values as for $k \in \mathbb N$, you have $f(2k \pi)=-2k \pi <0$ and $f(2k \pi + \pi)=2k \pi + \pi >0$. Consequently, $f$ is not injective and do not have an inverse.

Now considering the interval $[0,\pi]$, which I saw a bit late is the original question...

You have $$f^\prime(x)=\cos x -\cos x +x \sin x=x \sin x$$ which is stricly positive in $(0,\pi)$. So $f$ is strictly increasing on $(0,\pi)$. As $f$ is continuous, $f$ is a bijection from $(0, \pi)$ to $f[(0,\pi)]=(0,\pi)$. And as $f(0)=0$ and $f(\pi)=\pi$, $f$ is a bijection from $[0,\pi]$ to itself.

mathcounterexamples.net
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  • True. But in the question the function is considered only on $0\leq x \leq \pi$. And it is strictly increasing there... – bartgol Nov 30 '15 at 15:58
  • is it true that since f(x) is strictly increasing in that interval, its range is from 0 to pi and it has an inverse? – Be Happy Nov 30 '15 at 16:04
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    Yes $f$ has an inverse on that interval. Which doesn't mean that you can write the inverse in term of known functions. – mathcounterexamples.net Nov 30 '15 at 16:08
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As I mentioned in other comments, I believe you plugged the values in $f'$ instead of $f$. As a matter of fact,

$$ f(0) = 0,\qquad f(\pi) = \pi $$

Hence the maximum is $\pi$ and the minimum is 0.

For the inverse existence, a sufficient condition is that the function be strictly increasing or strictly decreasing on the whole domain. Can you prove that?

bartgol
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