giggles123
let f(x)=ax^3+6x^2+bx+4. Determine the constant a and b such that f has a relative min at x=1 and a rel max at x=2@Algebra



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Hero
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I hate these

moneybird
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first derivative of f(x) first

giggles123
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I did that then what?
I got F'(x)=3ax^2+12x+b

myininaya
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\[f'(x)=3a^2+12x+b\]
\[f'(1)=0 ;f'(2)=0\]

myininaya
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\[f'(x)=3ax^2+12x+b*\]

Hero
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I was just about to say....what happened to the x?

myininaya
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\[f'(1)=3a12+b=0\]

myininaya
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\[f'(2)=3a(4)+12(2)+b=0\]

myininaya
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\[3a+b=12\]
\[12a+b=24\]

moneybird
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3a + b =12
12a + 24 +b = 0
12a + b =24
(3a + b) = 12
9a = 36
a = 4
b =24

giggles123
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what happened to the x?

myininaya
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\[ 3a+b=12\]
\[(12a+b=24)\]

9a+0b=36
9a=36
a=4
12+b=12
b=24
gj money

myininaya
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what do you mean what happen to x?

giggles123
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how did you go from f'(x)=3ax^2+12x+b* to f′(−1)=3a−12+b=0

myininaya
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\[f(a)=4x^3+6x^2+24x+4\]

myininaya
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1 and 2 are critical numbers
this is a polynomial
so our critical numbers only exist when f'=0

myininaya
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f'(1)=0
f'(2)=0

myininaya
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this was given to us

myininaya
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money and i applied these conditions to f'

giggles123
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I got it. Thank you so much!

luckey
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here b can take any value and a is rrestricted to any value in the interval (2,2)