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

1. 3/6 or 1/22. 1/123. 1/84. 1/105. 2/12 or 1/6