(tangent function behavior lc)\nthe function f(x) = tan x changes from a decreasing rate of change to an…

Answer:

We first find the second - derivative of (y = f(x)=\tan x).

The first - derivative of (y=\tan x) is (y'=\sec^{2}x).

The second - derivative of (y) with respect to (x) is (y'' = 2\sec x\cdot\sec x\tan x=2\sec^{2}x\tan x).

We want to find where the function changes from a decreasing rate of change to an increasing rate of change. This occurs at the inflection points, where (y'' = 0) or is undefined.

Set (y''=2\sec^{2}x\tan x = 0). Since (\sec^{2}x=\frac{1}{\cos^{2}x}\neq0) for all real (x) such that (\cos x\neq0), we solve (\tan x = 0).

The solutions of (\tan x = 0) are (x = n\pi), (n\in\mathbb{Z}).

We are interested in the interval ((\frac{3\pi}{2},\frac{5\pi}{2})). When (n = 2), (x = 2\pi) and (2\pi\in(\frac{3\pi}{2},\frac{5\pi}{2})).

So the function (y = \tan x) changes from a decreasing rate of change to an increasing rate of change at (x = 2\pi) in the given interval.

Explanation:

Step1: Find first - derivative

(y'=\sec^{2}x)

Step2: Find second - derivative

(y'' = 2\sec^{2}x\tan x)

Step3: Set second - derivative equal to 0

(2\sec^{2}x\tan x=0)

Step4: Solve for (x) considering the interval

Since (\sec^{2}x\neq0), solve (\tan x = 0). In the interval ((\frac{3\pi}{2},\frac{5\pi}{2})), (x = 2\pi)