The function f(x) = tan(4x) has vertical asymptotes wherever the tangent function is undefined. This occurs at odd multiples of π/2. To find the specific asymptotes for the function over the interval from x = 0 to x = 2, we first need to set the equation:
4x = (2n + 1)π/2
where n is any integer. Dividing both sides by 4 gives:
x = (2n + 1)π/8
Now we will determine the valid values of n such that x remains within the interval [0, 2].
- For n = 0: x = (2(0) + 1)π/8 = π/8 ≈ 0.3927 (valid as 0 < 2)
- For n = 1: x = (2(1) + 1)π/8 = 3π/8 ≈ 1.1781 (valid as 1.1781 < 2)
- For n = 2: x = (2(2) + 1)π/8 = 5π/8 ≈ 1.9635 (valid as 1.9635 < 2)
- For n = 3: x = (2(3) + 1)π/8 = 7π/8 ≈ 2.7489 (invalid as 2.7489 > 2)
Thus, the valid asymptotes for the function f(x) = tan(4x) within the range of x = 0 to x = 2 are located at:
- x = π/8 (approximately 0.3927)
- x = 3π/8 (approximately 1.1781)
- x = 5π/8 (approximately 1.9635)