Find the absoluto maximum and minimum, il either exists, for the function on the indicated interval. \( I(x)=x^{3}-15 x^{2}+63 x+4 \) \( \begin{array}{lll}\text { (A) }[-1,10] & \text { (B) }[-1,7] & \text { (C) }[6,10]\end{array} \) (A) Find the absolute maximum. Select the correct choice below and, if necessary fill in the answer boxes to complete your choice. A. The absolute maximum is \( \square \) at \( x=\square \) (Use a comma to soparate answers as needed.) B. There is no absolute maximum.

To find the absolute maximum and minimum of the function \( I(x) = x^{3} - 15x^{2} + 63x + 4 \) on the interval \([-1, 10]\), we need to evaluate the function at critical points and endpoints. First, we find the derivative: \( I'(x) = 3x^{2} - 30x + 63 \). Setting \( I'(x) = 0 \): \( 3x^{2} - 30x + 63 = 0 \) Dividing by 3 gives: \( x^{2} - 10x + 21 = 0 \). Factoring the quadratic: \( (x - 3)(x - 7) = 0 \) yields critical points at \( x = 3 \) and \( x = 7 \). Now evaluate \( I \) at the endpoints and critical points: - \( I(-1) = (-1)^{3} - 15(-1)^{2} + 63(-1) + 4 = -1 - 15 - 63 + 4 = -75 \) - \( I(3) = 3^{3} - 15(3)^{2} + 63(3) + 4 = 27 - 135 + 189 + 4 = 85 \) - \( I(7) = 7^{3} - 15(7)^{2} + 63(7) + 4 = 343 - 735 + 441 + 4 = 53 \) - \( I(10) = 10^{3} - 15(10)^{2} + 63(10) + 4 = 1000 - 1500 + 630 + 4 = 134 \) Now, the function values are: - At \( x = -1 \), \( I(-1) = -75 \) - At \( x = 3 \), \( I(3) = 85 \) - At \( x = 7 \), \( I(7) = 53 \) - At \( x = 10 \), \( I(10) = 134 \) The absolute maximum is \( 134 \) at \( x = 10 \), and the absolute minimum is \( -75 \) at \( x = -1 \). A. The absolute maximum is \( 134 \) at \( x=10 \).