calculus
posted by Cam on .
Let f(x)=ax^3+6x^2+bx+4. Determine the constants a and b so that f has a relative minimu at x=1 and a relative maximumat x=2.

f ' (x) = 3ax^2 + 12x + b
given:
f ' (1) = 0 > 3a 12 + b = 0 or 3a + b = 12
f ' (2) = 0 > 12a + 24 + b = 0 or 12a + b = 24
subtract them:
9a = 36
a = 4
back in 3a + b = 12
12 + b = 12
b = 24
a = 4, b = 24
check:
f ' (x) = 12x^2 + 12x + 24
= 0 for a max/min
x^2  x  2 = 0
(x2)(x+1)=0
x = 2 or x = 1 , yeahhhh!