Calculus (help steve)
posted by Bob on .
f(x)=−8x^3+6ax^2−3bx+4 has a local minimum at x=1 and a local maximum at x=3. If a and b are the local minimum and maximum, respectively, what is the absolute value of a+b?

f(x) = −8x^3+6ax^2−3bx+4
f'(x) = 24x^2 + 12ax  3b
Now, local extrema occur where f'(x) = 0.
Since we see that f(x) is quadratic, and we know two zeros, we know that those are the only two zeros.
So, we know that
f'(x) = k(x1)(x3)
= kx^2  4kx + 3k
Take the antiderivative to see that
f(x) = k/3 x^3  2kx^2 + 3kx + C
Equating coefficients with the given f(x), we have
k/3 = 8, so k = 24
2k = 48 = 6a, so a = 8
3k = 72 = 3b, so b = 24
a+b = 32
Check:
f(x) = 8x^3 + 48x^2  72x + 4
f'(x) = 24x^2 + 96x  72
= 24(x^2  4x + 3)
= 24(x1)(x3) 
Thanks Steve.
But apparently, that's not the right answer, since a and b are not local minimum and maximum values. 
Got me. f(x) has a and b as coefficients.
Don't see how they can also be the local min and max.
Maybe you can figure out what they're asking.
Hmmm.
f(x) = −8x^3+6ax^2−3bx+4
f(1) = 8+6a3b+4 = 4+6a3b
f(3) = 216+54a9b+4 = 212+54a9b
So, if a and b are min and max, then
4+6a3b=a
212+54a9b=b
or
5a3b = 4
54a10b = 212
a = 149/28
b = 211/28
a+b = 90/7
Check:
f(x) = 8x^3 + 447/14 x^2  633/28 x + 4
f(1) = 149/28
f(3) = 211/28
If that's not right, I'm stumped what they want. 
It's not right either. I know, I'm stumped too.

But thanks for all the help.

i had trouble with this problem too

Could it be that the coefficients a and b are not the min and max a and b?
If f(x) = 8x^3 + 48x^2  72x + 4
Then f(1) = 28 and f(3) = 4
In that case, min+max (or, a+b) = 24 
Oops. a+b = 24