March 10th, 2008, 06:20 PM  #1 
Member Joined: Feb 2008 From: home Posts: 30 Thanks: 0  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  
Member Joined: Feb 2008 From: home Posts: 30 Thanks: 0  Quote:
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  
Member Joined: Feb 2008 From: home Posts: 30 Thanks: 0  Quote:
 

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