find the linearization ( l(x) ) of ( f(x)=\tan x ) at ( x=\frac{3 pi}{4} ).\nthe linearization is given by (…

Explanation:

Step1: Recall the formula for linearization

The formula for the linearization (L(x)) of a function (y = f(x)) at (x = a) is (L(x)=f(a)+f^{\prime}(a)(x - a)).

Step2: Find (f(a))

Given (f(x)=\tan x) and (a=\frac{3\pi}{4}). Then (f\left(\frac{3\pi}{4}\right)=\tan\left(\frac{3\pi}{4}\right)=- 1).

Step3: Find (f^{\prime}(x)) and (f^{\prime}(a))

Since (f(x)=\tan x), then (f^{\prime}(x)=\sec^{2}x=\frac{1}{\cos^{2}x}). Substitute (x = \frac{3\pi}{4}) into (f^{\prime}(x)), we have (f^{\prime}\left(\frac{3\pi}{4}\right)=\frac{1}{\cos^{2}\left(\frac{3\pi}{4}\right)}). Because (\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}), then (\cos^{2}\left(\frac{3\pi}{4}\right)=\frac{1}{2}), and (f^{\prime}\left(\frac{3\pi}{4}\right)=2).

Step4: Substitute into the linearization formula

Substitute (f(a)=-1), (f^{\prime}(a) = 2) and (a=\frac{3\pi}{4}) into (L(x)=f(a)+f^{\prime}(a)(x - a)). (L(x)=-1+2\left(x-\frac{3\pi}{4}\right)). Expand the expression: (L(x)=2x-\frac{3\pi}{2}-1).

Answer:

(L(x)=2x - 1-\frac{3\pi}{2})