Calculate the coordinaties of the poinc for which the congent 0 f \( f(x)=x^{3}-3 x+5 x+4 \) is perpendicular to the ling \( y=-\frac{1}{14} x+1 \) \( f^{\prime}(x)=3 x^{2}-6 x+5 \)

To find the coordinates of the point where the tangent to the function \( f(x) = x^3 - 3x + 5x + 4 \) is perpendicular to the line \( y = -\frac{1}{14}x + 1 \), we first note that the slope of the line is \(-\frac{1}{14}\). For the slopes to be perpendicular, the product of the slopes must equal \(-1\). Therefore, we want to find \( m_{tangent} \) such that: \[ m_{tangent} \cdot (-\frac{1}{14}) = -1 \implies m_{tangent} = 14. \] Next, we set the derivative \( f^{\prime}(x) = 3x^2 - 6x + 5 \) equal to 14: \[ 3x^2 - 6x + 5 = 14 \implies 3x^2 - 6x - 9 = 0. \] Dividing through by 3 gives \( x^2 - 2x - 3 = 0 \). Factor this: \[ (x - 3)(x + 1) = 0. \] Thus, \( x = 3 \) or \( x = -1 \). Now, we calculate \( f(x) \) at these values: 1. For \( x = 3 \): \[ f(3) = 3^3 - 3(3) + 5(3) + 4 = 27 - 9 + 15 + 4 = 37. \] So, one point is \( (3, 37) \). 2. For \( x = -1 \): \[ f(-1) = (-1)^3 - 3(-1) + 5(-1) + 4 = -1 + 3 - 5 + 4 = 1. \] The other point is \( (-1, 1) \). The coordinates of the points are \( (3, 37) \) and \( (-1, 1) \).