Question

Let $f\left(x\right)=x^3+p{x}^2+qx$?
a) Find the values of $p$ and $q$ so that $f(-1)=-8$ and $f^{\prime }(-1)=12$
b) Find the value of $p$ so that graph of $f$ changes concavity at $x=2$
c) Under what conditions of $p$ and $q$ will the graph of $f$ be increasing everywhere?

Answer 1

a) For the first condition to be true,
$-8=(-1{)}^3+p(-1{)}^2+q(-1)\to -8=-1+p-q\to -7=p-q$
The derivative with respect to $x$ is $f^{\prime }\left(x\right)=3x^2+2px+q$.

For condition two to be valid,

$12=3(-1{)}^2+2(-1)p+q\to 12=3-2p+q\to 9=q-2p$
So we have a system of equations.
$\{\begin{array}{c}p-q=-7\\ q-2p=9\end{array}$
Solving through elimination

we get $-p=2\to p=-2$

which means that $q=5$.
b) Concavity is determined by the second derivative.
$f^{\prime }\left(x\right)=3x^2+2px+q$
$f^{\prime \prime }\left(x\right)=6x+2p$
For $f\left(x\right)$ to change concavity $f^{\prime \prime }\left(x\right)=0$.
$0=6x+2p$
$0=6\left(2\right)+2p$
$-12=2p$
$p=-6$
c) For a function to be increasing on all $x,f^{\prime }\left(x\right)>0$ on all $x$.
$f^{\prime }\left(x\right)=3x^2+2px+q$
$0<3x^2+2px+q$
I don't know what you can do to simplify this further.