Register to reply 
Caculus: Critical Numbers 
Share this thread: 
#1
Jan2408, 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. 


#2
Jan2408, 10:22 PM

P: 1,754

Let's clear up any confusion about what the problem is ...
[tex]f'(x)=\frac{\cos^2 x}{x}\frac 1 5[/tex] Correct? 


#3
Jan2408, 10:40 PM

P: 12

Correct



#4
Jan2508, 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.



#5
Jan2508, 06:22 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,490

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= 5cos^{2}(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. 


Register to reply 
Related Discussions  
Critical numbers  Calculus  5  
Another critical numbers problem  Calculus & Beyond Homework  4  
A critical numbers problem  Calculus & Beyond Homework  15  
Critical Numbers  Calculus  1  
Critical numbers  Calculus & Beyond Homework  2 