where are the asymptotes of f(x) = tan(2x - π) from x = π/2 to x = 3π/2?\no a x = 3π/4, x = 5π/4\no b x = 0…

Explanation:

Step1: Recall tangent - asymptote formula

The tangent function $y = \tan(u)$ has asymptotes at $u=(2n + 1)\frac{\pi}{2}$, where $n$ is an integer. For the function $y=\tan(2x-\pi)$, we set $2x-\pi=(2n + 1)\frac{\pi}{2}$.

Step2: Solve for $x$

First, add $\pi$ to both sides of the equation $2x-\pi=(2n + 1)\frac{\pi}{2}$: [2x=(2n + 1)\frac{\pi}{2}+\pi=\frac{(2n + 1)\pi+2\pi}{2}=\frac{(2n + 3)\pi}{2}] Then divide both sides by 2: $x=\frac{(2n + 3)\pi}{4}$.

Step3: Find $n$ values for the given interval

We want to find the asymptotes in the interval $\frac{\pi}{2}\leq x\leq\frac{3\pi}{2}$. When $n = 0$, $x=\frac{3\pi}{4}$. When $n = 1$, $x=\frac{(2\times1 + 3)\pi}{4}=\frac{5\pi}{4}$.

Answer:

A. $x=\frac{3\pi}{4},x = \frac{5\pi}{4}$