Calculus help f(x) = a x^3 + b x^2 + c x + d?

-64a+16b-4c+d=4 , 125a+25b+5c+d=-4 , f'=3ax^2+2bx+c=0 , 48a-8b+c=0 , 75a+10b+c=0 , solve eqns we get a=16/729 , b= -8/243 , c= -320/243 , d=484/729

Take the derivative:

f'(x) = 3ax² + bx + c

This should have zeros for x = -4 and x = 5:

f'(x) = A * (x + 4)(x - 5)

Now you can multiply out then integrate...but that will yield a difficult form to deal with...instead, use a u-substitution:

f'(x) = A * u * (x - 5)

-->

u = x + 4 --> du = dx

x = u - 4 --> x - 5 = (u - 4 - 5) = u - 9

-->

f'(u) = A * u * (u - 9)

--> now you multiply out and it will be an easy form to integrate:

f'(u) = A * (u² - 9u)

--> if we make A really equal A = 6 * C, then we can make sure we get integers when we integrate (since we need integers divisible by both 3 and 2):

f'(u) = A * (6u² - 54u)

(this A is different from the original)

f(u) = A * (2u³ - 27u² + C)

--> but this is really just...notice A * C = D, the leftover constant

f(x) = A * (2(x + 4)³ - 27(x + 4)²) + D

Now evaluating @ x = -4 is easy:

f(-4) = 0 * A + D = 4 --> D = 4

Now we can evaluate @ x = 5, so find A:

using x + 4 --> 5 + 4 = 9

f(5) = A * (2 * 9³ - 27 * 9²) + D = -4

-->

-4 = -729A + 4

-->

A = 8/729

So the final answer is:

f(x) = 8/729(2(x + 4)³ - 27(x + 4)²) + 5

Which if you want to look like the above, you have to multiply out:

which is just going to introduce more ugly looking fractions, but you can do it if you keep everything straight...multiply the inside (what's multiplied by 8/729) then collect all like terms...then finally, distribute the 8/729

Ok - we have four constraints here:

f(-4) = 4

f(5) = -4

f'(-4) = 0

f'(5) = 0

First, let's calculate f'(x):

f'(x) = 3ax^2 + 2bx + c

So these equations become:

-64a + 16b - 4c + d = 4

125a + 25b + 5c + d = -4

48a - 8b + c = 0

75a + 10b + c = 0

You need to solve this system of 4 equations and four variables. Yuck.

Let's try this a simpler way. Since f'(-4) = f'(5) = 0, we can write:

f'(x) = a(x+4)(x-5) = a(x^2 - x - 20)

Now integrating, we get:

f(x) = a/3 x^3 - a/2 x^2 - 20ax + C

Now we have two equations and two unknowns - much easier.

f(-4) = -64/3 a - 8a + 80a + C = 4

152/3 a + C = 4

f(5) = 125/3 a - 25/2 a - 100a + C = -4

(250 - 75 - 600)/6 a + C = -4

-425/6 a + C = -4

(152/3 + 425/6) a = 8

729/6 a = 8

243/2 a = 8

a = 16/243

You can get the other coefficients from there.

First, let's get the two points taken care of.

-64a + 16b - 4c + d = 4

125a + 25b + 5c + d = -4

f '(x) = 3ax² + 2bx + c = 0

48a - 8b + c = 0

75a + 10b + c = 0

Now solve for a,b,c and d simultaneously.