November 29th, 2011, 10:02 AM  #1 
Newbie Joined: Nov 2011 Posts: 27 Thanks: 0  Critical points and min/max again.
Looking for help on these 3 problems (out of 10 problems total): Find the critical points y = (5)/(sin x + 4) on [0,2pi] Find the critical points f(x)=4e^(x)  16e^(2x) Local min x= Local max x= Find the critical points y= ln(x^11) + ln((4x^2)^11), (0,2) Thanks 
November 29th, 2011, 10:19 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  Re: Critical points and min/max again.
sin(x) + 4 is not equal to zero for any real x, so f(x) is continuous, and differentiable, except on the endpoints of the interval [0, 2?]. d/dx(5)/(sin x + 4) = 5cos(x)/(sin(x) + 4)². 5cos(x) = 0 for x = ?/2, x = 3?/2 on [0, 2?], so the critical points are (?/2, 1), (3?/2, 5/3). 
November 29th, 2011, 10:47 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Critical points and min/max again.
1.) The denominator cannot be zero for real values of x, thus we need only look at: On [0,2?] we find the critical values are 2.) Multiply through by : =3\ln(2)" /> To determine the nature of the extremum, we can: i) look at the slope on either side of the critical value. f(x) is increasing. f(x) is decreasing. Thus is a local/global maximum. There is no local minimum, as: ii) look at concavity at the critical value. f(x) is concave down at the critical value, thus it is a local maximum. 3.) The zeroes of the denominator are excluded from the domain, so we need only consider: We take only the positive root, as the negative is excluded from the domain. 
November 29th, 2011, 10:59 AM  #4 
Newbie Joined: Nov 2011 Posts: 27 Thanks: 0  Re: Critical points and min/max again.
Thanks greg and mark, you once again have powered through what I considered impossible. Thanks.


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