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I've been working on this exercise for a while but I couldn't find any way out of it. A formula for the derivative of a function $f$ is given and I'm asked to find how many critical points it has.

$$f'(x) = 1 + \frac{210\sin(x)}{x^2-6x+10}$$

Here is what I've done:

$$f'(x) = 1 + \frac{210\sin(x)}{x^2-6x+10} = \frac{(x-3)^2+1+210\sin(x)}{(x-3)^2+1}$$

Now, my critical points are going to be where my $f'(x)$ is either equal to zero or not defined. My $f'(x)$ is always defined since the denominator $(x-3)^2+1=0$ has no zeros. However $f'(x)$ will be equal to zero everytime numerator $(x-3)^2+1+210\sin(x)=0$ I graphed this function and I found it has 10 zeros, so from this I concluded that my function has 10 critical points.

Is there any way to prove this analytically? How can I find the critical points of $f'(x)$?


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You can certainly show that are only finitely many solutions by showing that for large values of $|x|$, we have $\left| \frac{(x-3)^2 + 1}{210} \right| > 1$. This occurs for $x \in (- \infty, 3 - \sqrt{209}) \cup (3 + \sqrt{209}, \infty)$. Note that $\sqrt{209} < \sqrt{225} = 15$, so we need only worry about $x \in (-12,18)$. Furthermore, you know exactly when $\sin(x)$ is increasing/decreasing, and you know exactly where $\sin(x)$ is positive/negative, so you can determine the exact number of solutions from this information. – JavaMan Jan 21 '12 at 1:47
The critical points can only be approximated numerically. You can formally verify that the information from the graphing program is correct by appealing to the Intermediate Value Theorem and convexity. Unfortunately that will be $10$ similar but separate arguments. – André Nicolas Jan 21 '12 at 2:36
up vote 2 down vote accepted

Essentially you are looking for solutions to $(x-3)^2+1 = -210\sin(x)$.

Since $(x-3)^2+1 \gt 210$ if $|x-3| \gt \sqrt{209} \approx 14.457$, you only need consider intersections inside the range $[-11.457,17.457]$ which might be written as about $[-3.65\pi,5.56\pi]$. So you might expect about $9.2$ solutions.

You can count them: there is a pair around $x=-3.5 \pi$, a pair around $-1.5 \pi$, and $0.5 \pi$, $2.5 \pi$ and $4.5 \pi$. So there are five pairs of solutions making ten individual solutions to $(x-3)^2+1 = -210\sin(x)$ and to $f'(x)=0$.

Hence there are ten critical points of $f$.

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