Need Help with coefficients of g(x) = ax^3 + bx^2 + cx + d

[mace519]

g(x)= ax^3 + bx^2 + cx + d ; it has a local max at (2,4), local min at (0,0), find a b c d

I get g`(x)= 3ax^2 + 2bx + c
sub x=2 and x=0
c=0 and a= -b/3

not sure what to do after that.

 

[soroban]

Re: Need Help

Hello, mace519!

$\displaystyle g(x)\:=\: ax^3\,+\, bx^2\,+\,cx\,+\,d$
It has a local max at $\displaystyle (2,4)$, local min at $\displaystyle (0,0)$.
Find $\displaystyle a,\;b,\;c,\;d$

I get: $\displaystyle g'(x)\:=\:3ax^2\,+\,2bx\,+\,c$
sub $\displaystyle x = 2$ and $\displaystyle x = 0$
$\displaystyle c\,=\,0$ and $\displaystyle a\,=\,-\frac{b}{3}\;$ . . . Good!

Don't forgot . . . $\displaystyle (0,0)$ and $\displaystyle (2,4)$ are on the curve, too.

So $\displaystyle g(0)\,=\,0:\;\;a\cdot0^3\,+\,b\cdot0^2\,+\,c\cdot0\,+\,d\:=\:0\;\;\Rightarrow\;\;d\,=\,0$

And $\displaystyle \,f(2)\,=\,4:\;\;a\cdot2^3\,+\,b\cdot2^2\,+\,c\cdot2\,+\,d\:=\:4$

Since $\displaystyle c = 0$ and $\displaystyle d = 0$, we have: $\displaystyle \,8a\,+\,4b\:=\:4$

With this equation and your $\displaystyle a\,=\,-\frac{b}{3}$, you can solve for $\displaystyle a$ and $\displaystyle b$.