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November 29th, 2011, 11:02 AM   #1
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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((4-x^2)^11), (0,2)

Thanks
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November 29th, 2011, 11:19 AM   #2
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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).

-5cos(x) = 0 for x = ?/2, x = 3?/2 on [0, 2?], so the critical points are (?/2, 1), (3?/2, 5/3).
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November 29th, 2011, 11:47 AM   #3
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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.

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November 29th, 2011, 11:59 AM   #4
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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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