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

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Homework Help Overview

The problem involves determining the number of critical numbers for the function f based on its first derivative, f '(x) = (cos²(x)/x) - 1/5, within the interval (0,10). The original poster expresses confusion regarding the discrepancy between their graphical observations and the expected number of critical points.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster graphed the derivative and identified 6 critical points, questioning why the expected number is 3. Other participants clarify the definition of critical points and suggest focusing on the original derivative rather than differentiating again.

Discussion Status

The discussion is active, with participants clarifying the nature of critical points and the original poster seeking further explanation for the observed discrepancy in critical numbers. There is no explicit consensus yet, but guidance has been offered regarding the correct approach to identifying critical points.

Contextual Notes

Participants note that the derivative does not exist at x=0, which is relevant to the analysis within the specified interval. The original poster's previous incorrect answer is acknowledged, but the reasons for the confusion remain under discussion.

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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 ...

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

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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