Question

find the linearization L(x) of f(x)=tanx at x=pi

Answer 1

f' = sec^2(Pi/4) = 2,so tan(Pi/4) = 1,

and, L(x) = 2(x - Pi/4) + 1

Answer 2

The 'linearization' name's a fancy way of saying you're getting the tangent at the point x = Pi/4. So, L(x) = f' (Pi/4)(x - Pi/4) + f(Pi/4)