# Calculus (help steve)

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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?

• Calculus (help steve) - ,

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(x-1)(x-3)
= 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(x-1)(x-3)

• Calculus (help steve) - ,

Thanks Steve.
But apparently, that's not the right answer, since a and b are not local minimum and maximum values.

• Calculus (help steve) - ,

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+6a-3b+4 = -4+6a-3b
f(3) = -216+54a-9b+4 = -212+54a-9b

So, if a and b are min and max, then

-4+6a-3b=a
-212+54a-9b=b
or
5a-3b = 4
54a-10b = 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.

• Calculus (help steve) - ,

It's not right either. I know, I'm stumped too.

• Calculus (help steve) - ,

But thanks for all the help.

• Calculus (help steve) - ,

i had trouble with this problem too

• Calculus (help steve) - ,

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

• Calculus (help steve) - ,

Oops. |a+b| = 24