Calculus and Analytic Geometry

The equation of the tangent line to a curve $y = f(x)$ at the point $(x_1, y_1)$ is given by the formula $y - y_1 = f'(x_1)(x - x_1)$. Finding the derivative of the given function, we get $f'(x) = 2x - 2$. Substituting the point $(1, 0)$, we get $f'(1) = 2(1) - 2 = 0$. Therefore, the equation of the tangent line is $y - 0 = 0(x - 1) = -x + 1$. Simplifying, we get $y = -x + 2$.

To find the equation of the tangent line, we need to find the slope of the curve at the point ((1, 0)). The slope is given by the derivative of the function, which is (f'(x) = 3x^2 - 4x + 3). Substituting (x = 1), we get (f'(1) = 3(1)^2 - 4(1) + 3 = 2). Therefore, the equation of the tangent line is (y - 0 = 2(x - 1)), which simplifies to (y = x - 1).

To find the equation of the tangent line, we need to find the slope of the curve at the point ((2, 1)). The slope is given by the derivative of the function, which is (f'(x) = x^2 - 4x + 4). Substituting (x = 2), we get (f'(2) = 2^2 - 4(2) + 4 = 0). Therefore, the equation of the tangent line is (y - 1 = 0(x - 2)), which simplifies to (y = 3x - 1).

To find the length of the curve (y = x^3 - 2x^2 + 3x - 4) from (x = 0) to (x = 2), we need to use the formula (L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx), where (\frac{dy}{dx}) is the derivative of the function that defines the curve and ([a, b]) is the interval of integration. In this case, (\frac{dy}{dx} = 3x^2 - 4x + 3) and ([a, b] = [0, 2]).

The ancient Indians are credited with developing the concept of calculus. They made significant contributions to the field, including the development of differentiation and integration.