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 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((4-x^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: 7,879 Thanks: 1087 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: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Critical points and min/max again. 1.) $y=\frac{5}{\sin x+4}=5$$\sin x+4$$^{\small{-1}}$ $\frac{dy}{dx}=-5$$\sin x+4$$^{\small{-2}}$$\cos x$$=-\frac{5\cos x}{$$\sin x+4$$^2}$ The denominator cannot be zero for real values of x, thus we need only look at: $\cos x=0$ On [0,2?] we find the critical values are $x=\frac{\pi}{2},\frac{3\pi}{2}$ 2.) $f(x)=4e^{-x}-16e^{-2x}$ $\frac{df}{dx}=-4e^{-x}+32e^{-2x}=0$ Multiply through by $\frac{e^{2x}}{4}$: $8-e^x=0$ $e^x=8$ $x=\ln(=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'(\ln(7)\)=-\frac{4}{7}+\frac{32}{49}=\frac{4}{49}>0$ f(x) is increasing. $f'(\ln(9)\)=-\frac{4}{9}+\frac{32}{81}=-\frac{4}{81}<0$ f(x) is decreasing. Thus $$$3\ln(2),f\(3\ln(2)$$\)=$$3\ln(2),\frac{1}{4}$$$ is a local/global maximum. There is no local minimum, as: $\lim_{x\to-\infty}=-\infty$ ii) look at concavity at the critical value. $f''(x)=4e^{-x}-64e^{-2x}$ $f''$$3\ln(2)$$=\frac{4}{8}-\frac{64}{64}=-\frac{1}{2}<0$ f(x) is concave down at the critical value, thus it is a local maximum. 3.) $y=\ln$$x^{11}$$+\ln$$\(4-x^2$$^{11}\)=11$$\ln(x)+\ln\(4-x^2$$\)=11\,\ln$$4x-x^3$$$ $\frac{dy}{dx}=\frac{4-3x^2}{4x-x^3}$ The zeroes of the denominator are excluded from the domain, so we need only consider: $4-3x^2=0$ We take only the positive root, as the negative is excluded from the domain. $x=\sqrt{\frac{4}{3}}$
 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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### Find the critical points of f(x) = 4eâˆ’x âˆ’ 16eâˆ’2x

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