I Critical points for difficult function

[Robin64]

Given that f'(x)=[(8cos(x)/(x²)-(⅛)], find the number of relative maxima and minima on the interval (1,10). Finding maxima and minima analytically wasn't fruitful for me, so instead I used a bit of handwaving. First I argued, using the intermediate value theorem, since f'(1)>f'(10), there exists at least one maximum or minimum. Then I said, well, the derivative is approximately equal to 8cos(x)/x² and since cos(x) has 2 minima and one maximum on the interval (1,10), then f(x) must have 2 minima and one maximum on that same interval.

Is there way to solve this without that handwaving?

 

[BvU]

Is there way to solve this without that handwaving?

Take the complete derivative and work it out ?

 

[Robin64]

Take the complete derivative and work it out ?

The derivative is the expression in the problem. Setting the derivative equal to zero results in 64cos(x)-x²=0. I don't see the way forward to solving that expression.

 

[BvU]

$$ f' (x) = 0 \Leftrightarrow ( 8 \sqrt{\cos x} + x )( 8 \sqrt{\cos x} - x ) = 0 \quad ? $$
[edit] Limits the search for extrema to $[1, {5\over 2} \pi] = [1, u]$ with $u < 8$

 

[Robin64]

Yep, I got that. That's where I'm stuck, solving 8sqrt(cos(x)+/-x=0. I thought about expressing sqrt(cos(x)) as the first few terms of a Maclaurin series but that's less than exact or elegant.

 

[BvU]

Exactness isn't required if you only need the number of extrema. A simultaneous plot of $\cos x$ and $x^2/64$ does the trick...

 

[Robin64]

I know. I was just unsatisfied with not being able to find an analytic solution.