Question

TT

Where are the asymptotes of f(x) = tan(2x - pi) from x = pi/2 to x = 3pi/2?
a. x = 3pi/4, x = 5pi/4
b. x = 0, x = 2pi
c. x = 0, x = pi/4
d. x = pi/2, x = 3pi/2

Answer 1

The vertical asymptotes of the function f(x) = tan(2x - π) from x = π/2 to x = 3π/2 are at x = 3π/4 and x = 5π/4. These are found by solving the equation for when the argument of the tangent function equals an odd multiple of π/2. Therefore correct option is A

Step-by-step explanation:

You asked about the asymptotes of the function f(x) = tan(2x − π) from x = π/2 to x = 3π/2. To find the vertical asymptotes of a tangent function, we look for the values of x where the argument of the tangent function,

in this case, 2x − π, is equal to an odd multiple of π/2 because the tangent function has undefined values at these points.

The general form for this is 2x − π = (2n+1)π/2, where n is an integer. Solving for x, we get

x = ((2n+1)π/2 + π)/2

= (2n+1)π/4 + π/2.

We need to find out what n gives us the values within the specified range.

For n = 1, we get x = 3π/4, and for n = 2, we get x = 5π/4, which are both within the given interval.

Thus, the correct answer is:

a. x = 3π/4, x = 5π/4