Questions: Mean value theorem h(z)=4z^3-8z^2+7z-2 on (2,5).

Solution Steps

The function \( h(z) = 4z^{3} - 8z^{2} + 7z - 2 \) is a polynomial, which is continuous and differentiable everywhere. Therefore, it satisfies the conditions of the Mean Value Theorem on the interval \([2, 5]\).

The derivative of the function is given by: \[ h'(z) = 12z^{2} - 16z + 7 \]

Next, we calculate the values of the function at the endpoints: \[ h(2) = 12 \] \[ h(5) = 333 \]

Using the Mean Value Theorem, we find: \[ mvt\_value = \frac{h(5) - h(2)}{5 - 2} = \frac{333 - 12}{3} = 107 \]

We set \( h'(c) = 107 \) and solve for \( c \): \[ 12c^{2} - 16c + 7 = 107 \] This simplifies to: \[ 12c^{2} - 16c - 100 = 0 \] Using the quadratic formula, we find: \[ c = \frac{16 \pm \sqrt{(-16)^{2} - 4 \cdot 12 \cdot (-100)}}{2 \cdot 12} = \frac{16 \pm \sqrt{256 + 4800}}{24} = \frac{16 \pm \sqrt{5056}}{24} \] This results in two potential solutions: \[ c = \frac{2}{3} - \frac{\sqrt{79}}{3} \quad \text{and} \quad c = \frac{2}{3} + \frac{\sqrt{79}}{3} \] Only the solution \( c = \frac{2}{3} + \frac{\sqrt{79}}{3} \) lies within the interval \( (2, 5) \).

Final Answer