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. (answer: p=-2 and q=5)
b)find the value of p so that the graph of f changes concavity at x=2. (answer: p=-6)
C) UNDER WHAT CONDITIONS ON P AND Q WILL THE GRAPH OF F BE INCREASING EVERYWHERE? (I don't get this problem...)
thanks!

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.  (answer: p=-2 and q=5)
b)find the value of p so that the graph of f changes concavity at x=2.  (answer: p=-6)
C) UNDER WHAT CONDITIONS ON P AND Q WILL THE GRAPH OF F BE INCREASING EVERYWHERE?   (I don't get this problem...)
thanks!

anonymous · Answer

take the first derivative and set it to greater than zero.  you'll have an inequality in terms of p and q.

anonymous · Answer

so I get $$3x ^{2} + 2px + q \ge 0$$

anonymous · Answer

I got that but I don't know how to solve that.