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 March 10th, 2008, 06:20 PM #1 Member   Joined: Feb 2008 From: home Posts: 30 Thanks: 1 relative min or max or neither can u do this step by step please............... Decide if the given value of x is a critical number of f, and if so, decide whether the point for x on f is a relative minimum, relative maximum, or neither. f(x) = 3(x)^4 - 4(x)^3 - 12(x)^2 + 24; x = 0 March 10th, 2008, 06:37 PM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Taking the first derivative of function f gives us f'(x) = 12x^3 - 12x^2 - 24x. Plug in x = 0 in the given equation above f'(0) = 0. Thus, x = 0 is one of the critical numbers of function f (since f'(x) = 12x(x^2 - x - 2) = 12x(x - 2)(x + 1), there are three critical numbers, hence not just one critical numbers). Taking the derivative of the derivative function gives us f"(x) = 36x^2 - 24x - 24. Test concavity or possible of inflection point by plug in x = 0 f"(0) = -24. This tells us that at x = 0, graph f is concave down, hence we have an relative maximum. March 10th, 2008, 07:06 PM   #3
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 Originally Posted by johnny Taking the first derivative of function f gives us f'(x) = 12x^3 - 12x^2 - 24x. Plug in x = 0 in the given equation above f'(0) = 0. Thus, x = 0 is one of the critical numbers of function f (since f'(x) = 12x(x^2 - x - 2) = 12x(x - 2)(x + 1), there are three critical numbers, hence not just one critical numbers). Taking the derivative of the derivative function gives us f"(x) = 36x^2 - 24x - 24. Test concavity or possible of inflection point by plug in x = 0 f"(0) = -24. This tells us that at x = 0, graph f is concave down, hence we have an relative maximum.

Thank you and can u just explain one more thing:
2(x)^3 - 3(x)^2 - 12x + 18 ; x = 2
f'(x)= 6(x)^2 - 6(x) - 12=> 6(x^2 - x - 2)=>6(x - 2)(x + 1)
here are 2 critical numbers x = 2 and x = 1

f'(2)= 24 - 12 - 12 => 0

what is relative minimum or maximium or neither? March 10th, 2008, 08:26 PM #4 Member   Joined: Feb 2008 Posts: 89 Thanks: 0 [color=darkblue]Hi: x^2 - x - 2 does not factor as (x - 2)(x - 1). Regards, Rich B. rmath4u2@aol.com [/color] March 11th, 2008, 04:42 PM   #5
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 Originally Posted by nikkor180 [color=darkblue]Hi: x^2 - x - 2 does not factor as (x - 2)(x - 1). Regards, Rich B. rmath4u2@aol.com [/color]
thank you, was my mistake supose to be X^- x - 2 = (x - 2)(x + 1) SOrry!! Tags max, min, relative Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Conway51 Applied Math 53 April 22nd, 2016 06:40 AM Joolz Computer Science 0 October 1st, 2012 12:38 PM echilon Computer Science 1 June 19th, 2012 08:51 AM mikeportnoy Algebra 1 February 18th, 2010 02:08 PM echilon Algebra 0 December 31st, 1969 04:00 PM

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