Why Does f'(x) Have Only 3 Critical Numbers on (0,10)?

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SUMMARY

The function's first derivative is given by f'(x) = (cos²(x)/x) - 1/5. The critical numbers of the function f on the interval (0,10) are determined by finding where f'(x) equals zero or does not exist. The correct number of critical points is 3, as the equation cos²(x) = 5x has three intersection points when graphed. The confusion arises from mistakenly analyzing the second derivative instead of the first derivative to identify critical points.

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Homework Statement


The first derivative of the function f is given by f '(x)= (((cos^2)x)/(x)) - 1/5
How many critical numbers does f have on the open interval (0,10)?


Homework Equations





The Attempt at a Solution


I already got this question wrong but I don't know why I got it wrong. The answer is 3 but when I graph it I see 6 critical numbers. So why is it 3 and not 6? Please explain.
 
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Let's clear up any confusion about what the problem is ...

f'(x)=\frac{\cos^2 x}{x}-\frac 1 5

Correct?
 
Correct
 
I tried taking the derivative of the derivative and graphed that but it still gives me an image of 6 critical points.
 
I don't understand why that would happen. A critical point occurs where the derivative is 0 or does not exist. Clearly your derivative does not exist at x= 0. To determine where f'= 0, I graphed y= 5cos2(x) and y= x. They cross in 3 points.

Wait, did you differentiate again? You said what you gave was f '. To determine where f has critical points, you should be graphing that, not its derivative.
 
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