SOLUTION: good day please help me to solve this problem If i have this function : f( x ) =1+ 12x – x^3 which is the minimun relative?

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Question 638252: good day please help me to solve this problem
If i have this function :
f( x ) =1+ 12x – x^3
which is the minimun relative?
Found 2 solutions by Tatiana_Stebko, stanbon:
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
the derivative f'( x )=(1+ 12x – x^3)'=12-3x^2
12-3x%5E2=0
x=2 or x=-2
f%282%29=1%2B+12%2A2+%96+2%5E3=17, f%28-2%29=1%2B+12%2A%28-2+%29%96+%28-2%29%5E3=-15
(2, 17) and (-2,-15) are the critical points
Let use Second derivative test
the second derivative f''( x )=(12-3x^2)=-6x
at critical point (2, 17) we find f''(2)=-6%2A2=-12%3C0, so (2, 17) is a relative maximum

at critical point (-2,-15) we find f''(-2)=-6%2A%28-2%29=12%3E0, so (-2,-15) is a relative minimum

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If i have this function :
f( x ) =1+ 12x – x^3
which is the minimun relative?
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Take the derivative.
f'(x) = -3x^2+12
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Solve:
-3x^2 + 12 = 0
x^2 = 4
x = +2 or x = -2
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At this point you might use the 2nd devivative test.
f''(x) = -6x
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f(-2) = 12, so relative min at x = -2
f(2) = -12, so relative max at x = 2
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graph%28400%2C400%2C-10%2C10%2C-25%2C25%2C1%2B12x-x%5E3%29
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Cheers,
Stan H.
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