To approximate the number e-0.03 using a tangent line, it makes sense to... O Find the linearization of f (x) = e" at a - -0.03 and evaluate it at x - -0.03. O Find the linearization of f(x) = er at a - 0 and evaluate it at x - 0. O Find the linearization of f(x) = et at a = -0.03 and evaluate it at x - 0. O Find the linearization of f(a) = et at a = 0 and evaluate it at x - -0.03