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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.​

1 Answer

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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}

User Ralan
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