# I Critical points for difficult function

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1. Mar 30, 2017

### 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?

2. Mar 30, 2017

### BvU

Take the complete derivative and work it out ?

3. Mar 30, 2017

### Robin64

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.

4. Mar 30, 2017

### BvU

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

Last edited: Mar 30, 2017
5. Mar 30, 2017

### 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.

6. Mar 30, 2017

### 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...

7. Mar 30, 2017

### Robin64

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