Final answer:
To find the equation of the tangent line to the curve at the point (1, 4), we need to find the derivative of the curve and evaluate it at x = 1. The derivative of the function f(x) = 5x² - x³ is f'(x) = 10x - 3x². Evaluating this derivative at x = 1 gives us the slope of the tangent line, which is 7. Using the point-slope form of a linear equation, we can find the equation of the tangent line: y = 7x - 3.
Step-by-step explanation:
The equation of a tangent line to a curve can be found by calculating the derivative of the curve at the given point. In this case, the curve is defined by the function f(x) = 5x² - x³ and the point of tangency is (1, 4). To find the slope of the tangent line, we need to find the derivative of f(x) and evaluate it at x = 1. The slope of the tangent line is equal to the derivative of f(x) at x = 1. The derivative of f(x) is found using the power rule, which states that the derivative of x^n is equal to n*x^(n-1).
Using the power rule, we find that the derivative of f(x) = 5x² - x³ is f'(x) = 10x - 3x². Evaluating this derivative at x = 1 gives us the slope of the tangent line at the point (1, 4). Therefore, the slope of the tangent line is f'(1) = 10*1 - 3*1² = 7.
Now that we have the slope of the tangent line and the point of tangency, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. Plugging in the values for the point (1, 4) and the slope 7, we get y - 4 = 7(x - 1). Simplifying this equation gives us the equation of the tangent line: y = 7x - 3.