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March 10th, 2008, 06:20 PM   #1
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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
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March 10th, 2008, 06:37 PM   #2
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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.
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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?
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March 10th, 2008, 08:26 PM   #4
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[color=darkblue]Hi:

x^2 - x - 2 does not factor as (x - 2)(x - 1).

Regards,

Rich B.
rmath4u2@aol.com [/color]
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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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