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Question 1101261: find values for a,b,c,and d so that the function f (x) =ax³ + bx² + cx + d has a relative maximum at (-2,6) and a relative minimum at (2,-10)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! f (x) =ax³ + bx² + cx + d
f(0)=d
use point (-2, 6)
f(-2)=-8a+4b-2c+d=6
f(2)=8a+4b+2c+d=-10
add these and 8b+2d=-4, or 4b+d=-2, d=-4b-2
Now the derivatives are 3ax^2+2bx+c=0, and that occurs when x=-2 and x=2. So we have
12a-4b+c=0
12a+4b+c=0
24a+2c=0
12a+c=0, c=-12a b=0 and d=-2
Therefore, -8a-2c-2=6 and -8a-2c=8 or -4a-c=4
and 8a+2c-2=-10 and 8a+2c=-8 or 4a+c=-4 (the same thing as above)
c=-12a
4a-12a=-4 and -8a=-4 a=1/2
c=-6
f(x)=(1/2)x^3-6x-2
a=1/2
b=0
c=-6
d=-2
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