My Math Forum Relative Maxima and Minima

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 March 24th, 2011, 10:56 AM #1 Newbie   Joined: Mar 2011 Posts: 17 Thanks: 0 Relative Maxima and Minima Hey everyone, I've got a question that I have no clue how to answer: Let f(x)= 4 - 6/x + 4/x^2. Find the open intervals on which f is increasing (decreasing. Then determine the x-coordinates of all relative maxima (minima). 1. f is increasing on the intervals: 2. f is decreasing on the intervals: 3. The relative maxima of occur at = 4. The relative minima of occur at = If someone could please explain what I am looking for here, it would be greatly appreciated. I gave myself the permission to skip the last 2 math classes in order to do essays, and now, as a result, I have no clue what's going on in math . Even the PDF files the teacher posted online make no sense to me lol.
 March 24th, 2011, 11:44 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Relative Maxima and Minima We can find possible extrema by setting the first derivative of the function equal to zero, and solving for x, to find the function's critical values (critical values also occur where f'(x) does not exist, but these values are typically at vertical asymptotes, not relative extrema). If the sign of f'(x) changes across a critical value, then there is an extremum there. f(x) is increasing if f'(x) > 0 and decreasing if f'(x) < 0. One way to conclude the extremum is a maximum is if f'(x) is positive before the extremum and negative after the extremum, and if the reverse is true then we conclude the extremum is a minimum. Another way is by checking the sign of f''(x) at the extremum point. If f''(x) is negative, then the extremum is a maximum, and if f''(x) is positive then the extremum is a minimum. You have: $f(x)=4-6x^{-1}+4x^{-2}$ $f'(x)=6x^{-2}-8x^{-3}=\frac{2$$3x-4$$}{x^3}$ We see $f'$$\frac{4}{3}$$=0$ and $f'(0)$ does not exist. So we check the intervals: $$$-\infty,0$$$ we can use x = -1: f'(-1) is -/- = + so f(x) is increasing $$$0,\frac{4}{3}$$$ we can use x = 1: f'(1) is -/+ = - so f(x) is decreasing $$$\frac{4}{3},\infty$$$ we can use x = 2: f'(2) is +/+ = + so f(x) is increasing We can then conclude that a relative minimum occurs at $x=\frac{4}{3}$.
 March 24th, 2011, 11:55 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Relative Maxima and Minima f(x) = 4 - 6/x + 4/x^2, f'(x) = 6/x² - 8/x³ = (6x - 8 )/x³. 1. f is increasing on the intervals where its derivative is positive (upward slope). (-?, 0), (4/3, ?). 2. f is decreasing on the intervals where its derivative is negative (downward slope). (0, 4/3). 3. The relative maxima of f occur where the derivative is equal to zero and positive to the (slightly) left of this point and negative to the (slightly) right of this point. None! 4. The relative minima of f occur where the derivative is equal to zero and negative to the (slightly) left of this point and positive to the (slightly) right of this point. x = 4/3. Saw your post, Mark, but since this is the second time I've been preposted today I'm going to send my blurb out there anyway.
 March 24th, 2011, 12:02 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Relative Maxima and Minima It certainly doesn't hurt to have more than one way to look at a problem!
 March 24th, 2011, 03:00 PM #5 Newbie   Joined: Mar 2011 Posts: 17 Thanks: 0 Re: Relative Maxima and Minima Thanks guys, both of you were very helpful
 March 24th, 2011, 03:07 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Relative Maxima and Minima You're welcome! And welcome to the forum too!
 March 24th, 2011, 04:58 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Welcome!
 March 24th, 2011, 07:46 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Welcome to the forum, MFP.
 March 25th, 2011, 06:43 AM #9 Newbie   Joined: Mar 2011 Posts: 17 Thanks: 0 Re: Relative Maxima and Minima Thanks guys

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# let f(x)=8âˆ’4/x 8/x2. find the open intervals on which f is increasing (decreasing). then determine the x-coordinates of all relative maxima (minima).

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