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!