(ii) Find the value of \( f^{\prime}(x) \) when \( x=1 \). \[ f(x)=6(1)^{2}-6(1)^{1}+7=6-6+^{\prime} \] (iii) Find the equation of the tangent to the curve \( y=f(x) \) at \( x=1 \). (A,

Certainly! Let's tackle each part of your question step by step. --- ### Given Function: \[ f(x) = 6x^2 - 6x + 7 \] --- ### (ii) Finding \( f^{\prime}(x) \) at \( x = 1 \) **Step 1: Find the derivative \( f^{\prime}(x) \).** To find the derivative of \( f(x) \) with respect to \( x \), apply the power rule to each term: \[ f(x) = 6x^2 - 6x + 7 \] \[ f^{\prime}(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(7) \] \[ f^{\prime}(x) = 12x - 6 + 0 \] \[ f^{\prime}(x) = 12x - 6 \] **Step 2: Evaluate \( f^{\prime}(x) \) at \( x = 1 \).** \[ f^{\prime}(1) = 12(1) - 6 = 12 - 6 = 6 \] **Answer:** \[ f^{\prime}(1) = 6 \] --- ### (iii) Finding the Equation of the Tangent to the Curve \( y = f(x) \) at \( x = 1 \) To find the equation of the tangent line at a specific point on the curve, you need two pieces of information: 1. The **slope** of the tangent line at that point. 2. A **point** through which the tangent line passes. **Step 1: Determine the slope of the tangent line.** From part (ii), we have already found that: \[ f^{\prime}(1) = 6 \] So, the slope \( m \) of the tangent line at \( x = 1 \) is **6**. **Step 2: Find the point on the curve at \( x = 1 \).** Calculate \( f(1) \): \[ f(1) = 6(1)^2 - 6(1) + 7 = 6 - 6 + 7 = 7 \] So, the point of tangency is \( (1, 7) \). **Step 3: Use the point-slope form to write the equation of the tangent line.** The point-slope form of a line is: \[ y - y_1 = m(x - x_1) \] Where: - \( m \) is the slope. - \( (x_1, y_1) \) is the point on the line. Plugging in the known values: \[ y - 7 = 6(x - 1) \] **Simplify the equation:** \[ y - 7 = 6x - 6 \] \[ y = 6x - 6 + 7 \] \[ y = 6x + 1 \] **Answer:** \[ \text{The equation of the tangent line is } y = 6x + 1. \] --- **Summary:** - **(ii)** \( f^{\prime}(1) = 6 \) - **(iii)** Equation of the tangent at \( x = 1 \) is \( y = 6x + 1 \)