anonymous
  • anonymous
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
Calculus1
chestercat
  • chestercat
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Hero
  • Hero
I hate these
anonymous
  • anonymous
first derivative of f(x) first
anonymous
  • anonymous
I did that then what? I got F'(x)=3ax^2+12x+b

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myininaya
  • myininaya
\[f'(x)=3a^2+12x+b\] \[f'(-1)=0 ;f'(2)=0\]
myininaya
  • myininaya
\[f'(x)=3ax^2+12x+b*\]
Hero
  • Hero
I was just about to say....what happened to the x?
myininaya
  • myininaya
\[f'(-1)=3a-12+b=0\]
myininaya
  • myininaya
\[f'(2)=3a(4)+12(2)+b=0\]
myininaya
  • myininaya
\[3a+b=12\] \[12a+b=-24\]
anonymous
  • anonymous
3a + b =12 12a + 24 +b = 0 12a + b =-24 -(3a + b) = 12 9a = -36 a = -4 b =24
anonymous
  • anonymous
what happened to the x?
myininaya
  • myininaya
\[ 3a+b=12\] \[-(12a+b=-24)\] ------------------ -9a+0b=36 -9a=36 a=-4 -12+b=12 b=24 gj money
myininaya
  • myininaya
what do you mean what happen to x?
anonymous
  • anonymous
how did you go from f'(x)=3ax^2+12x+b* to f′(−1)=3a−12+b=0
myininaya
  • myininaya
\[f(a)=-4x^3+6x^2+24x+4\]
myininaya
  • myininaya
-1 and 2 are critical numbers this is a polynomial so our critical numbers only exist when f'=0
myininaya
  • myininaya
f'(-1)=0 f'(2)=0
myininaya
  • myininaya
this was given to us
myininaya
  • myininaya
money and i applied these conditions to f'
anonymous
  • anonymous
I got it. Thank you so much!
anonymous
  • anonymous
here b can take any value and a is rrestricted to any value in the interval (-2,2)

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