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Question
. Let f be the 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. Determine 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]
Since f(x) = x³ + ax² + bx + c, it is known that f'(x) = 3x² + 2ax + b and that f"(x) = 6x + 2a.
Since the inflection point is at x=0, this means f"(0)=0 at this point, which is 6(0) + 2a = 0.
Thus, a = 0, so the function is really f(x) = x³ + bx + c.
The average on the interval [0,2] would be[F(2) - F(0)]/(2-0).
It can be seen that F(x) = x^4/4 + bx^2/2 + cx + K.
If we compute F(2) - F(0) we get 4 + 4b + 2c.
When this is divided by 2-0, which is 2, the answer is 2 + 2b + c, and that's the average.
Looking back at the function, at x=0, the function has the value -2, so this means c=-2.
This means our average 2 + 2b + c is really just 2b.
To be -3, 2b = -3, so b = -1.5.
Putting this altogether gives us a function of x³ - 1.5x - 2.
To satisfy the mean value theorem on [0,3], compute the integral over this interval,
divide by 3, and that is the average value. Once this is done, find where the function
is equal to that value.
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