I have this homework question that asks.
Find the values of a, b, c and d, such that g(x) = ax^3 +bx^2 +cx +d has a local max at (2,4) and a local min at (0,0)
Its really confusing me in how to approach it.
Use the equation and its derivative to build a system of equations. It should be rather simplistic.
The max and min occur at (0,0) and (2,4)
We have:
$\displaystyle a(0)^{3}+b(0)^{2}+c(0)+d=0$, therefore, d=0.
$\displaystyle a(2)^{3}+b(2)^{2}+c(2)+d=4$
Derivatives must be 0 at extrema:
$\displaystyle 3a(2)^{2}+2b(2)+c=0$
$\displaystyle 3a(0)^{2}+2b(0)+c=0$, therefore, c=0.
Because c and d are 0, that helps whittle things down a good bit.
We have the following system to solve:
$\displaystyle 8a+4b=4$
$\displaystyle 12a+4b=0$
We find a=-1 and b=3
Therefore, the polynomial is $\displaystyle g(x)=-x^{3}+3x^{2}$