Question

NEED HELP ASAP!!!

Let f(x) = ax^3 + bx^2 +cx+d. Determine the values of a, b, c, and d so that f(x) critical
points at x=2 and x=-1, F(0) = 1, and F'(0) = 6.​

Answer 1

`f(x)=ax^3+bx^2+cx+d`

has derivative

`f'(x)=3ax^2+2bx+c`

We want this to have critical points at
`x=2` and
`x=-1`; we want
`f'` to be 0 or undefined at these points. Since
`f` is a polynomial, it is continuous, so the derivative will always exist.

So we need

`\begin{cases}12a+4b+c=0\\3a-2b+c=0\end{cases}`

We also want
`f(0)=1` and
`f'(0)=6`. This means

`\begin{cases}d=1\\c=6\end{cases}`

so the previous system reduces to

`\begin{cases}6a+2b=-3\\3a-2b=-6\end{cases}\implies a=-1,b=\frac32`

So the function is

`\boxed{f(x)=-x^3+\frac32x^2+6x+1}`