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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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Quote:
 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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Quote:
 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!!

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