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How I can prove this?

Let $$f(x) = \sin x + \frac{x^2}{2} - 2x.$$

Show that $f$ has only one critical number in the interval $[2, 3]$.

It belongs to the chapter on applications of the derivative. Section: Relative extrema of real functions.

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up vote 2 down vote accepted

The definition of "critical point" is that it is a point where the function is defined, but the derivative is either equal to $0$ or undefined.

So, let's try taking the derivative and staring at it, perhaps we can "tell" immediately that there is one and only one critical point in $[2,3]$.

$$\begin{align*} f(x) &= \sin x + \frac{x^2}{2} -2x\\ f'(x) &= \cos x + x - 2. \end{align*}$$ This is defined everywhere, so the only critical points will be stationary points.

So... can we just "tell" that $f'(x)$ takes the value $0$ only once on $[2,3]$ by staring at it? Hmmm... That doesn't seem easy to do (you could note that $x-2$ takes the values from $0$ to $1$, and that $\cos x$ is negative and decreasing, but that tells you that $f'(x)$ is negative at $2$ and positive at $3$, hence $f$ has at least one critical point by the Intermediate Value Theorem since $f'$ is continuous, but it doesn't tell you it has exactly one).

What else can we do? How about seeing whether this function is strictly monotone? As noted parenthetically, $f'(2)\lt 0$ and $f'(3)\gt 0$. If $f'(x)$ is strictly monotone on $[2,3]$, then it goes through zero exactly once.

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A very clear answer. And to prove that $f'$ is strictly increasing enough to calculate $f''$ and verify that is positive. Thanks so much. Served me well. =) – mathsalomon Aug 19 '11 at 4:29
@mathsalomon: If you found Arturo's answer helpful, you can vote it up by clicking on the up arrow. You can vote up to 30 times per day. Also you can do this on other questions when you read an answer which is particularly interesting or helpful. (you can even up vote questions!) – Eric Naslund Aug 19 '11 at 4:52
@Eric Naslund: Thanks Eric. – mathsalomon Aug 19 '11 at 7:17
@mathsalomon: Note that you can only "accept" one answer (I mention it because you seem to have switched two or more times between my answer and Eric's). That's the "checkmark" on the left, under the vote counter. You can change your mind, but only one answer can be the "accepted answer" at a time. If you are unsure which answer was, in your mind, the most helpful or clearest, then wait a while before accepting. – Arturo Magidin Aug 19 '11 at 15:04

Hint: The second derivative is positive on $[2,3]$. To see why, notice $$f''(x) =-\sin x+1>0$$ and that $|\sin x |<1$ on $[2,3]$.

This means that the derivative $f^'(x)$ is increasing on $[2,3]$, and you can use this idea to find that there is at most one critical number. The intermediate value theorem tells you there is exactly one.

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I think you mean "negative"? – Arturo Magidin Aug 19 '11 at 3:36
He means $-\sin x + 1$ is positive. – Robert Israel Aug 19 '11 at 3:38
@Arturo: Thanks, corrected. – Eric Naslund Aug 19 '11 at 3:38
@Robert: Exactly. The way I had worded it earlier was very strange so I fixed it now. – Eric Naslund Aug 19 '11 at 3:41
Good answer. Thanks. – mathsalomon Aug 19 '11 at 4:32

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