Questions: Given the function of (f(x)=-3 x^2+7 x+8) on the interval ([-2,5]), find a point (c) between -2 and 5 so that it satisfies the Mean Value Theorem.

Solution Steps

To find a point \( c \) that satisfies the Mean Value Theorem (MVT) for the function \( f(x) = -3x^2 + 7x + 8 \) on the interval \([-2, 5]\), we need to:

1. Ensure the function is continuous and differentiable on the given interval.
2. Calculate the average rate of change of the function over the interval.
3. Find the value of \( c \) in the interval \([-2, 5]\) where the derivative of the function equals the average rate of change.

The function \( f(x) = -3x^2 + 7x + 8 \) is a polynomial, which is continuous and differentiable on the interval \([-2, 5]\). Therefore, the conditions for the Mean Value Theorem are satisfied.

We compute the values of the function at the endpoints: \[ f(-2) = -3(-2)^2 + 7(-2) + 8 = -18 \] \[ f(5) = -3(5)^2 + 7(5) + 8 = -32 \] The average rate of change over the interval \([-2, 5]\) is given by: \[ \text{Average Rate of Change} = \frac{f(5) - f(-2)}{5 - (-2)} = \frac{-32 - (-18)}{7} = \frac{-14}{7} = -2 \]

Next, we find the derivative of the function: \[ f'(x) = -6x + 7 \] Setting the derivative equal to the average rate of change: \[ -6c + 7 = -2 \] Solving for \( c \): \[ -6c = -2 - 7 \implies -6c = -9 \implies c = \frac{9}{6} = \frac{3}{2} \]

The value \( c = \frac{3}{2} \) is within the interval \([-2, 5]\).

Final Answer