Find the coordinates of points where the graph of \( f(x)=-x^{3}+9 x^{2}-15 x+11 \) has horizontal tangents. \( (x, y)=(\square) \) (smaller \( x \)-value) \( (x, y)=(\square \)-value)

To find the coordinates of the points where the graph of \( f(x) = -x^{3} + 9x^{2} - 15x + 11 \) has horizontal tangents, we first need to locate the derivative \( f'(x) \) and set it equal to zero. Calculating the derivative: \[ f'(x) = -3x^{2} + 18x - 15 \] Setting \( f'(x) = 0 \): \[ -3x^{2} + 18x - 15 = 0 \] Dividing by -3 results in: \[ x^{2} - 6x + 5 = 0 \] Factoring: \[ (x - 5)(x - 1) = 0 \] Thus, the \( x \)-values where the horizontal tangents occur are: \[ x = 1 \quad \text{and} \quad x = 5 \] Now, substituting these \( x \)-values back into \( f(x) \) to find their corresponding \( y \)-coordinates. For \( x = 1 \): \[ f(1) = -(1)^{3} + 9(1)^{2} - 15(1) + 11 = -1 + 9 - 15 + 11 = 4 \] For \( x = 5 \): \[ f(5) = -(5)^{3} + 9(5)^{2} - 15(5) + 11 = -125 + 225 - 75 + 11 = 36 \] Therefore, the points where the graph has horizontal tangents are: \( (x, y) = (1, 4) \) (smaller \( x \)-value) \( (x, y) = (5, 36) \) (larger \( x \)-value)