If f ' (x) = 3x^2 + 1, find the equation of the tangent line to f (x) = x^3 + x at x = - 1.

F (x) = x³ + x → f ' (x) = 3x² + 1

f ' (-1) = 3 + 1 = 4 represents m, the slope at x = - 1.

y = mx + b ↔ y = 4x + b

for x = - 1 f (x) = - 1³ - 1, so y = - 2

-2 = 4 (-1) + b

-2 = - 4 + b and b = 2

Then the equation of te tangent line at x = - 1 is y = 4x+2