Question

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 ?

Answer 1

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

Answer 2

first equation gives us p-q=9
second gives -2p+q=9

Answer 3

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

Answer 4

so f(x)=x^3+-18x^2+-27x
f'(x)=3x^2-18*2x-27=3x^2-36x-27
f''(x)=6x-36

Answer 5

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

Answer 6

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

Answer 7

remember you use the second derivative to find inflection points so thats why i found f''

Answer 8

thanks you very much! do you know how to do part c?

Answer 9

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}

Answer 10

oops my inequality signs the first one is suppose to say greater
second one less than k?

Answer 11

we want p(-q+3p)>0

Answer 12

I got this from inside the square root thing up above

Answer 13

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

Answer 14

okay.