2. Find the slop of the lure and the equation of the tangent line to the function \[ f(x)=x^{3}-3 x^{2}+4 x-2 \quad \text { at point when } x=1 \]

To find the **slope of the curve** and the **equation of the tangent line** to the function \( f(x) = x^{3} - 3x^{2} + 4x - 2 \) at the point where \( x = 1 \), follow these steps: ### 1. Find the Slope of the Curve at \( x = 1 \) The **slope of the curve** at a particular point is given by the derivative of the function at that point. **Step 1:** Compute the derivative \( f'(x) \). \[ f(x) = x^{3} - 3x^{2} + 4x - 2 \] \[ f'(x) = \frac{d}{dx}(x^{3}) - \frac{d}{dx}(3x^{2}) + \frac{d}{dx}(4x) - \frac{d}{dx}(2) \] \[ f'(x) = 3x^{2} - 6x + 4 \] **Step 2:** Evaluate the derivative at \( x = 1 \). \[ f'(1) = 3(1)^{2} - 6(1) + 4 = 3 - 6 + 4 = 1 \] So, the **slope of the curve** at \( x = 1 \) is **1**. ### 2. Find the Equation of the Tangent Line at \( x = 1 \) The equation of the tangent line can be found using the point-slope form of the equation of a line: \[ y - y_{1} = m(x - x_{1}) \] Where: - \( m \) is the slope of the tangent line. - \( (x_{1}, y_{1}) \) is the point of tangency. **Step 1:** Find the \( y \)-coordinate of the point where \( x = 1 \). \[ f(1) = (1)^{3} - 3(1)^{2} + 4(1) - 2 = 1 - 3 + 4 - 2 = 0 \] So, the point of tangency is \( (1, 0) \). **Step 2:** Use the point-slope form with \( m = 1 \) and \( (x_{1}, y_{1}) = (1, 0) \). \[ y - 0 = 1(x - 1) \] \[ y = x - 1 \] ### **Final Answer:** - **Slope of the curve at \( x = 1 \):** \( 1 \) - **Equation of the tangent line:** \( y = x - 1 \)