Final answer:
To find f'(1), we take the derivative of f(x) = 9x² - x³ and evaluate it at x = 1. The derivative is f'(x) = 18x - 3x². Evaluating at x = 1, we get f'(1) = 15. The equation of the tangent line to y = 9x² - x³ at (1, 8) is y = 15x - 7.
Step-by-step explanation:
To find f'(1), we need to take the derivative of the function f(x) = 9x² - x³ and evaluate it at x = 1. The derivative of a function is found by applying the power rule and the chain rule. So, the derivative of f(x) is f'(x) = 18x - 3x². Evaluating at x = 1, we get f'(1) = 18(1) - 3(1)² = 18 - 3 = 15.
To find the equation of the tangent line to the curve y = 9x² - x³ at the point (1, 8), we can use the point-slope form of a line. The point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is the point on the curve and m is the slope of the tangent line. Plugging in the values, we have y - 8 = 15(x - 1). Simplifying, we get y = 15x - 7.