Linearization of $f(x)=-\cos(x)+1-\lambda$

Question

What is the linearization of $$ f(x)=-\cos(x)+1-\lambda, $$ around $x_0$, where $\lambda$ is a constant?

I think its $$ f(x)=f(x_0)+f'(x_0)(x-x_0)=-\cos(x_0)+1-\lambda+\sin(x_0)(x-x_0) $$

In particular, if $f(x_0)=0$, then the linearization around $x=x_0$ is $$ f(x)=\sin(x_0)(x-x_0). $$

Answer 1

Yes it is correct but, to be rigorous, we should write

$$f(x)=f(x_0)+f'(x_0)(x-x_0)+o(x-x_0)=\\=-\cos(x_0)+1-\lambda+\sin(x_0)(x-x_0)+o(x-x_0)$$

and then the linearization is

$$g(x)=-\cos(x_0)+1-\lambda+\sin(x_0)(x-x_0)$$

Answer 2

In general if $f$ is a differentiable function on some open interval $(a,b)$ then the equation $g(x)$ of the tangent line at the point $(x_0,f(x_0))$ is given by $$g(x)=g(x_0)+f'(x_0)(x-x_0)=f(x_0)+f'(x_0)(x-x_0)$$ where $f'(x_0)$ is the derivative of $f$ at $x_0$ and of course $f(x_0)=g(x_0)$ since $f$ and $g$ touch each other at $x_0$ (you can even draw a picture to convince yourself). With $f(x)=-\cos x+1-\lambda$ $$g(x)=g(x_0)+\sin x_0(x-x_0)=-\cos x_0+1-\lambda+\sin x_0(x-x_0)$$