anonymous
  • anonymous
let f(x)= x^3 + px^2 +qx a) find the values of p and q so that f(-1) =8 and f'(-1)=12 b) find the value of p so that f has a point of inflection at x=2 (i.e. f''(2)=0) c) under what conditions of p and q will df/dx > 0 for all x ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
f(x)=x^3+px^2+qx f(-1)=-1+p-q=8 f'(x)=3x^2+2px+q f'(-1)=3-2p+q=12 so we have -1+p-q=8 and 3-2p+q=12
myininaya
  • myininaya
first equation gives us p-q=9 second gives -2p+q=9
myininaya
  • myininaya
first equation gives p=9+q second+first gives -2(9+q)+q=9 =>-18-2q+q=9 =>-18-q=9 => -18-9=q so q=-27 then p=9+q=9+(-27)=-18

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myininaya
  • myininaya
so f(x)=x^3+-18x^2+-27x f'(x)=3x^2-18*2x-27=3x^2-36x-27 f''(x)=6x-36
myininaya
  • myininaya
we don't really need to find the second derivate yet since we are suppose to find new p and q but above is a unless you see a mistake somewhere
myininaya
  • myininaya
f(x)=x^3+px^2+qx f'(x)=3x^2+2px+q f''(x)=6x+2p f''(2)=6(2)+2p=0 solve for p so we have 6(2)+2p=0 so 12+2p=0 so p=-6
myininaya
  • myininaya
remember you use the second derivative to find inflection points so thats why i found f''
anonymous
  • anonymous
thanks you very much! do you know how to do part c?
myininaya
  • myininaya
we have f(x)=x^3+px^2+qx so f'(x)=3x^2+2px+q we want f'(x)>0 so we want 3x^2+2px+q>0 3x^2+2px>-q x^2+2px/3>-q/3 (divided both sides by 3) x^2+2px/3+(2p/2*3)^2>-q/3+(2p/2*3)^2 (We want to complete the square) (x+p/3)^2>-q/3+p^2/9 so we have x+p/3>sqrt{-q/3+p^2/9} => x=-p/3+sqrt{-q/3+p^2/9} x+p/3<-sqrt{-q/3+p^2/9} => x=-p/3-sqrt{-q/3+p^2/9}
myininaya
  • myininaya
oops my inequality signs the first one is suppose to say greater second one less than k?
myininaya
  • myininaya
we want p(-q+3p)>0
myininaya
  • myininaya
I got this from inside the square root thing up above
myininaya
  • myininaya
i found a common denominator inside i got (-qp+3p^2)/(27) but we don't care about 27 since it is >0 -qp+3p^2>0 or p(-q+3p)>0
anonymous
  • anonymous
okay.

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