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anonymous
 5 years ago
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 ?
anonymous
 5 years ago
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 ?

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1f(x)=x^3+px^2+qx f(1)=1+pq=8 f'(x)=3x^2+2px+q f'(1)=32p+q=12 so we have 1+pq=8 and 32p+q=12

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1first equation gives us pq=9 second gives 2p+q=9

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1first equation gives p=9+q second+first gives 2(9+q)+q=9 =>182q+q=9 =>18q=9 => 189=q so q=27 then p=9+q=9+(27)=18

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1so f(x)=x^3+18x^2+27x f'(x)=3x^218*2x27=3x^236x27 f''(x)=6x36

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1we 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
 5 years ago
Best ResponseYou've already chosen the best response.1f(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
 5 years ago
Best ResponseYou've already chosen the best response.1remember you use the second derivative to find inflection points so thats why i found f''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks you very much! do you know how to do part c?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1we 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/3sqrt{q/3+p^2/9}

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1oops my inequality signs the first one is suppose to say greater second one less than k?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1I got this from inside the square root thing up above

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i 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
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