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I have a general idea about it, but I can't calculate it correctly ... Question: Find a cubic function, in the form below, that has a local maximum value of 3 at -4 and a local minimum value of 0 at 3. f (x) = ax3 + bx2 + cx + d What I am doing now: f(-4)=3---64a+16b-4c+d=3 f(3)=0---27a+9b+3c+d=0 f'(-4)=f'(3)=0---a=b f'(x)=3ax^2+2bx+c
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and I got c=-12a and d=84a a=196/3 I think they're wrong..
f (x) = ax^3 + bx^2 + cx + d
how do you get a=b ? o.O you should be getting 3a=2b.. please recheck..
yeeeahhh !!! I got it !!! damnnn.. I always make such stupid mistakes!!! thank you soooo MUCH !
glad to help ^_^
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