f(x) = x^3 - x at x = -1 we have the point (-1, 0)
The slope of the tangent line at any point on the curve can be found by taking the derivative of the function....so....
f ' (x) = 3x^2 - 1
And at -1, the slope = 3(-1)^2 - 1 = 3 - 1 = 2
So.....the equation of the tangent line is
y = 2(x - (-1) )
y = 2(x +1)
y = 2x + 2
Here's a graph: https://www.desmos.com/calculator/ikztawuevq
It's 5!Ya'll slow.
