Posted by Leanna on Wednesday, February 9, 2011 at 6:39am.
From a=24,b=36,
we find
f'(x)=0 has roots at x=-3 and -1, so that checks.
f"(x)=0 has a root at x=-2, so that checks too.
To solve B, we integrate
f(x):=4*x^3+24*x^2+36*x+k
from 0 to 1:
I=∫f(x)dx
=x^4+8*x^3+18*x^2+k*x
Evaluate I from 0 to 1 to give k+27
But since is given as I=32, so
I=32=k+27
Solve for k.
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