# Calculus Help (deciphering a function and MVT)

• Feb 5th 2009, 03:18 PM
alakaboom1
Calculus Help (deciphering a function and MVT)
This is the problem:

Let f be a function defined by f(x)= x3+ax2+bx+c and having the following properties:

I) The graph of f has a point of inflection at (0, -2).
II) The average (mean) value of f(x) on the closed interval [0,2] is -3

a) What are the values 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]

I got that a=0 by calculating the second derivative, and then plugging in the inflection point. But I would really appreciate some help with the rest of the problem. Thanks to anyone who can help me out!!
• Feb 5th 2009, 04:59 PM
Jester
Quote:

Originally Posted by alakaboom1
This is the problem:

Let f be a function defined by f(x)= x3+ax2+bx+c and having the following properties:

I) The graph of f has a point of inflection at (0, -2).
II) The average (mean) value of f(x) on the closed interval [0,2] is -3

a) What are the values 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]

I got that a=0 by calculating the second derivative, and then plugging in the inflection point. But I would really appreciate some help with the rest of the problem. Thanks to anyone who can help me out!!

You also know that the inflection point is located at $(0,-2)$ so

$f(0) = 0^3 + b\cdot 0 + c = -2$ so this give you $c=-2$.

Nest, the average of the function is $-3$

so

$\frac{1}{2} \int_0^2 \left( x^3 + bx -2 \right)\, dx = -3$

which gives $b$ and thus $f(x)$. For the last part find $x_0$ such that

$f'\left( x_0 \right) = \frac{f(3) - f(0)}{3}$.

I think you can handle it from here.
• Feb 5th 2009, 05:38 PM
alakaboom1
Quote:

Originally Posted by danny arrigo
You also know that the inflection point is located at $(0,-2)$ so

$f(0) = 0^3 + b\cdot 0 + c = -2$ so this give you $c=-2$.

Nest, the average of the function is $-3$

so

$\frac{1}{2} \int_0^2 \left( x^3 + bx -2 \right)\, dx = -3$

which gives $b$ and thus $f(x)$. For the last part find $x_0$ such that

$f'\left( x_0 \right) = \frac{f(3) - f(0)}{3}$.

I think you can handle it from here.

Yes, thank you very much.