This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — Here are the solutions for parts (a) and (b). Given the curve y = x^3 - 3x - 1 and the point (1, -3). Step 1: Find the first derivative of the curve to get the slope of the tangent. (dy)/(dx) = (d)/(dx)(x^3 - 3x - 1) = 3x^2 - 3 Step 2: Calculate the slope of the tangent at the point (1, -3). Substitute x=1 into the derivative: m_t = (dy)/(dx) |_x=1 = 3(1)^2 - 3 = 3 - 3 = 0 Part (a): Equation of the tangent Step 3: Use the point-slope form y - y_1 = m_t(x - x_1) with m_t = 0 and (x_1, y_1) = (1, -3). y - (-3) = 0(x - 1) y + 3 = 0 y = -3 The equation of the tangent is y = -3. Part (b): Equation of the normal Step 4: Calculate the slope of the normal. The slope of the normal m_n is the negative reciprocal of the slope of the tangent m_t. Since m_t = 0, the tangent line is horizontal. Therefore, the normal line will be vertical. A vertical line has an undefined slope and its equation is of the form x = c. Since the normal passes through (1, -3), its x-coordinate must be 1. The equation of the normal is x = 1. What's next?