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 points where the tangent line to the function \( f(x) = x^{3} - 3x^{2} + 5x + 4 \) is perpendicular to the line \( y = -\frac{1}{14}x + 1 \), follow these steps: ### 1. **Understand Perpendicular Slopes** Two lines are perpendicular if the product of their slopes is \(-1\). Given the line: \[ y = -\frac{1}{14}x + 1 \] - **Slope (\( m_1 \))**: \(-\frac{1}{14}\) Let the slope of the tangent line to \( f(x) \) be \( m_2 = f'(x) \). For the lines to be perpendicular: \[ m_1 \times m_2 = -1 \] \[ -\frac{1}{14} \times f'(x) = -1 \] \[ f'(x) = 14 \] ### 2. **Find the Derivative of \( f(x) \)** Given: \[ f(x) = x^{3} - 3x^{2} + 5x + 4 \] Differentiate \( f(x) \): \[ f'(x) = 3x^{2} - 6x + 5 \] ### 3. **Set the Derivative Equal to 14 and Solve for \( x \)** \[ 3x^{2} - 6x + 5 = 14 \] \[ 3x^{2} - 6x - 9 = 0 \] \[ x^{2} - 2x - 3 = 0 \] *(Divided the entire equation by 3 for simplicity)* Solve the quadratic equation: \[ x = \frac{2 \pm \sqrt{4 + 12}}{2} \] \[ x = \frac{2 \pm \sqrt{16}}{2} \] \[ x = \frac{2 \pm 4}{2} \] Thus: \[ x = 3 \quad \text{or} \quad x = -1 \] ### 4. **Find the Corresponding \( y \)-Coordinates** Plug these \( x \)-values back into \( f(x) \): - **For \( x = 3 \)**: \[ f(3) = 3^{3} - 3(3)^{2} + 5(3) + 4 = 27 - 27 + 15 + 4 = 19 \] \[ \text{Point: } (3, 19) \] - **For \( x = -1 \)**: \[ f(-1) = (-1)^{3} - 3(-1)^{2} + 5(-1) + 4 = -1 - 3 - 5 + 4 = -5 \] \[ \text{Point: } (-1, -5) \] ### 5. **Conclusion** The coordinates of the points where the tangent to \( f(x) \) is perpendicular to the given line are: - \( (3, \, 19) \) - \( (-1, \, -5) \)