# Calculus AB

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Find a, b, c, and d such that the cubic function ax^3 + bx^2 + cx + d satisfies the given conditions

Relative maximum: (2,4)
Relative minimum: (4,2)
Inflection point: (3,3)

So this is what I have so far:

f'(x) = 3a^2 + 2bx + c
f''(x) = 6ax + 2b

f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2

--> 30a + 6b + c = -1

Am I going in the right direction? If I am, what do I do next? Thanks.

• Calculus AB -

You have a good start, but there are a few corrections to make:

f'(x) = 3a(x)^2 + 2bx + c
f''(x) = 6ax + 2b

f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2

--> 30a + 6b + c = -1

You have three conditions for max. at (2,4):
f(2)=4
f'(2)=0
f"(2)<0

relative min. at f(4)=2 =>
f(4)=2
f'(4)=0
f"(4)>0

Inflexion point at (3,3) means:
f(3)=3
f"(3)=0

So you'd have six equations for four unknowns a,b, c and d. Notice that if the system is consistent (which is the case here), you only need 4 of the conditions to find the solutions, but you will have to check that the remaining conditions are satisfied.

Can you take it from here?

• Calculus AB -

your thinking is ok so far
also remember that (3,3) also lies on the original function, so
27a + 9b + 3c + d = 3 will be another equation.

How did you get 30a in 30a + 6b + c = -1
If you subtracted your two equations and then divided by 2 should it not have been
28a + 6b + c = -1 ? ---> #4

by subtracting my new equation from 64a +16b+4c+d = 2 , I get a new equation
37a + 7b+c=-1 ---> # 5

#5 - #4 :
9a + b = 0
b = -9a

also we know that f'(2) = 0 and f'(4) = 0
f'(2) = 12a + 4b + c = 0
f'(4) = 48a + 8b + c = 0
subtract:
36a + 4b = 0
b = -9a , nothing new here , confirms above.

but we also know f'(2) = f''(3) = 0
12a + 4b + c = 18a + 2b
c = 6a - 2b
c = 6a - 2(-9a) = 24a

in #4:
28a + 6(-9a) + 24a = -1
-2a = -1
a = 1/2
b= -9/2
c = 12

so now back into : 8a + 4b + 2c + d = 4
4 -18 +24 + d = 4
d = -6

so f(x) = (1/2)x^3 -(9/2)x^2 + 12x - 6

check my arithmetic

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