local max and min of g(x) = ax^3 + bx^2 + cx + d

[Vempy]

Find the values of a, b, c, and d, such that g(x) = ax^3+bx^2+cx+d has a local maximum at (2,4) and a local minimum at (0,0).

d must be zero because:
g(0)=a0^3+b0^2+c0+d
0=a0^3+b0^2+c0+d
d=0

g(2)=a(2)^3+b(2)^2+c(2)
4=8a+4b+2c
4=2(4a+2b+c)
2=4a+2b+c

I don't know where to go from here. I'm assuming doing derivatives comes into play somewhere in this.

 

[skeeter]

yes ... derivatives do come into play. extrema occur at critical values. critical values are where the derivative of a function equals 0 or is undefined. polynomial functions are well-behaved, and are differentiable everywhere.

you have a local max at (2,4) ... so, g'(2) = 0

you have a local min at (0,0) ... g'(0) = 0

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

since g'(0) = 0 ... c = 0

g'(2) = 12a + 4b = 0

you also had the original equation ...

2 = 4a + 2b + c ... and you know that c = 0, so 2 = 4a + 2b or 1 = 2a + b.

solve for a and b.