Answer
a. $L(x)=2x+\pi/2-1$.
See graph.
b. $L(x)=-\sqrt 2x-\sqrt 2\pi/4+\sqrt 2$.
See graph.
Work Step by Step
a. Given $f(x)=tan(x)$, we have $f'(x)=sec^2(x)$ and the linearization at $x=-\pi/4$ can be written as $L(x)=f(-\pi/4)+f'(-\pi/4)(x+\pi/4)=tan(-\pi/4)+sec^2(-\pi/4)(+\pi/4)=-1+2(x+\pi/4)=2x+\pi/2-1$.
See graph.
b. Given $g(x)=sec(x)$, we have $g'(x)=sec(x)tan(x)$ and the linearization at $x=-\pi/4$ can be written as $L(x)=g(-\pi/4)+g'(-\pi/4)(x+\pi/4)=sec(-\pi/4)+sec(-\pi/4)tan(-\pi/4)(x+\pi/4)=\sqrt 2+(\sqrt 2)(-1)(x+\pi/4)=-\sqrt 2x-\sqrt 2\pi/4+\sqrt 2$.
See graph.
