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January 31, 2015

January 31, 2015

Posted by **Savannah** on Friday, November 18, 2011 at 4:57pm.

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 -
**MathMate**, Friday, November 18, 2011 at 5:57pmYou have a good start, but there are a few corrections to make:

f'(x) = 3**a(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 -
**Reiny**, Friday, November 18, 2011 at 6:12pmyour 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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