Consider the curve \( f(x)=x-x^{3} \). (a) Find the slope of the tangent line to the curve at the point \( (1,0) \). (b) Find an equation of the tangent line in part (a). \( y=\square \)

To find the slope of the tangent line to the curve \( f(x) = x - x^3 \) at the point \( (1,0) \), we first compute the derivative of the function, \( f'(x) \). 1. Differentiate \( f(x) \): \[ f'(x) = 1 - 3x^2. \] 2. Evaluate the derivative at \( x = 1 \): \[ f'(1) = 1 - 3(1)^2 = 1 - 3 = -2. \] Thus, the slope of the tangent line at \( (1,0) \) is \( -2 \). Next, we find the equation of the tangent line using the point-slope form, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) = (1, 0) \). 1. Plugging in the values: \[ y - 0 = -2(x - 1). \] 2. Simplifying the equation: \[ y = -2x + 2. \] So the equation of the tangent line is: \[ y = -2x + 2. \]