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 \(\displaystyle b\).