Expert: Paul Klarreich - 3/7/2009

Question
Let F be the function by f(x)=x^3+ax^2+bx+c and having the following properties:

(i) The graph of f has an inflection point at (0,2)
(ii) the average mean value of f(x) on the closed interval [0,-2] is -3

a.) Determine the value of a, b, and c
b.) Determine the value of x that satisfies the conclusion of the mean value theorem for f on the closed interval [0,3]

Okay so I found out that both a and b equals -1 based on the properties, but I cant find out what c is and I'm stuck. Help please? I used the Average Mean equation and the second derivative to solve for a and b, but I cannot find c as it either cancels out or goes to zero. Did I overlook something?

Answer
Questioner:   Linrosa
Category:  Calculus
Private:  No
 
Subject:  Finding Values from points of inflection and avg. mean value
Question:  Let F be the function by f(x)=x^3+ax^2+bx+c and having the following properties:

(i) The graph of f has an inflection point at (0,2)
(ii) the average mean value of f(x) on the closed interval [0,-2] is -3

a.) Determine the value of a, b, and c
b.) Determine the value of x that satisfies the conclusion of the mean value theorem for f on the closed interval [0,3]

Okay so I found out that both a and b equals -1 based on the properties, but I cant find out what c is and I'm stuck. Help please? I used the Average Mean equation and the second derivative to solve for a and b, but I cannot find c as it either cancels out or goes to zero. Did I overlook something?
......................................

(i) f''(x) = 6x + 2a = 0 at x = 0.  So  a = 0.  (you got a = -1?)

f(0) = 2, so  c = 2.

(ii) The mean value of f(x) is  the INTEGRAL of f(x) over [-2,0], divided by 2.  (You do not write an interval with the larger number first.)

{
| (x^3 + bx + 2) dx = x^4/4 + bx^2/2 + 2x , from -2 to 0
}

= 16/4 + 2b - 4 = 2b.

Average value is 2b/2 = b, so b = -3.
........................................
Now use those to write your polynomial:  f(x) = x^3 - 3x + 2
and write:
       f(3) - f(0)
f'(c) = -----------
         3 - 0.

You should be able to handle that.