Caculus: Critical Numbers


by Hothot330
Tags: caculus, critical, numbers
Hothot330
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#1
Jan24-08, 10:19 PM
P: 12
1. The problem statement, all variables and given/known data
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)?


2. Relevant equations



3. 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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rocomath
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#2
Jan24-08, 10:22 PM
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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?
Hothot330
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#3
Jan24-08, 10:40 PM
P: 12
Correct

Hothot330
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#4
Jan25-08, 01:34 AM
P: 12

Caculus: Critical Numbers


I tried taking the derivative of the derivative and graphed that but it still gives me an image of 6 critical points.
HallsofIvy
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#5
Jan25-08, 06:22 AM
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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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