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 $f'(x)$?
Thanks!