Given the following function,
y=tan(pi-3x)
1)determine the interval for the principal cycle.
(^ Type your answer in interval notation)
2)Then for the principal cycle, determine the equations of the vertical asymptotes
3)find the coordinates of the center point
4) Find the coordinates of the halfway points.
( Direction - if needed type your points as ordered pair)
Thanks in advance. I'm clueless in this question. I have no idea where to begin or what to write.
Well, you know that the principal cycle for tan(x) is from
-π/2 < x < π/2
So, now you have
-π/2 < π-3x < π/2
-3π/2 < -3x < -π/2
π/2 > x > π/6
or, more usually written,
π/6 < x < π/2
But, you want the principal cycle to be near the origin, so shift it left one period (π/3) and you have
-π/6 < x < π/6
Use the same technique to answer the other questions. The graph is at
http://www.wolframalpha.com/input/?i=tan(%CF%80-3x)
1) The interval for the principal cycle can be determined by finding the period of the function. The period of the tangent function is π, so the interval for the principal cycle is (0, π).
2) The tangent function has vertical asymptotes at x = π/2 + kπ, where k is an integer. So, the equations of the vertical asymptotes for the principal cycle are x = π/2 and x = -π/2.
3) To find the coordinates of the center point, we need to find the average of the maximum and minimum values of the function over the principal cycle. Since the tangent function oscillates between positive and negative infinity, it does not have a center point. Instead, it has multiple vertical asymptotes.
4) The halfway points can be found by calculating the average of two consecutive vertical asymptotes. For the principal cycle, the halfway points are (π/2, 0) and (-π/2, 0).
No problem! Let's break down the given function step-by-step.
1) To determine the interval for the principal cycle, we need to consider the period of the tangent function. The period of the tangent function is π, which means it repeats every π units. In this case, we have a coefficient of 3 in front of x, which means the period is π/3. So, the interval for the principal cycle is (-∞, ∞).
2) Now, let's determine the equations of the vertical asymptotes. Vertical asymptotes occur when the denominator of the tangent function becomes zero. In this case, the denominator is cos(pi - 3x). To find when cosine is zero, we know that it happens at every multiple of π. So, setting pi - 3x = nπ, where n is an integer, we can solve for x.
pi - 3x = nπ
3x = pi - nπ
x = (pi - nπ)/3
Therefore, the equations of the vertical asymptotes are x = (pi - nπ)/3, where n is an integer.
3) To find the coordinates of the center point, we need to find the midpoint of the principal cycle. Since the interval for the principal cycle is (-∞, ∞), there is no specific center point in this case.
4) Lastly, let's find the coordinates of the halfway points. Since the period of the function is π/3, the halfway point occurs at x = π/6. Plugging this value into the function y = tan(pi - 3x), we can find the corresponding y-coordinate.
y = tan(pi - 3(π/6))
= tan(pi - π/2)
= tan(pi/2)
= undefined
Therefore, the coordinates of the halfway points are (π/6, undefined).
I hope this helps! Let me know if you have any further questions.
No worries! I'll guide you through step by step.
1) To determine the interval for the principal cycle of the function y = tan(pi-3x), we need to find the period of the tangent function. The period of the tangent function is π divided by the coefficient of x in the argument.
In this case, the coefficient of x is 3, so the period is π/3.
The principal cycle is the interval where the function repeats itself. Since the tangent function has a period of π/3, the principal cycle for this function is [-π/3, π/3]. Therefore, the interval for the principal cycle is (-π/3, π/3).
2) Next, let's identify the equations of the vertical asymptotes. Vertical asymptotes occur when the function has undefined values, which happens when the tangent function reaches values of ±π/2.
In the given function y = tan(pi-3x), we can find the vertical asymptotes by solving the equation pi-3x = ±π/2 for x.
For the first case, pi-3x = π/2, we solve for x:
pi-3x = π/2
3x = pi - π/2
3x = (2π - π)/2
3x = π/2
x = π/6
For the second case, pi-3x = -π/2:
pi-3x = -π/2
3x = pi + π/2
3x = (2π + π)/2
3x = 3π/2
x = π/2
So, the equations of the vertical asymptotes are x = π/6 and x = π/2.
3) Now, let's find the coordinates of the center point. The center point occurs at the middle of the principal cycle. Since the principal cycle is (-π/3, π/3), the center point is at x = 0 (the midpoint of the interval).
To find the y-coordinate of the center point, substitute x = 0 into the function y = tan(pi-3x):
y = tan(pi-3(0))
y = tan(pi) = 0
Therefore, the coordinates of the center point are (0, 0).
4) Finally, let's find the coordinates of the halfway points. The halfway points occur when the function reaches half of its maximum or minimum value within the principal cycle.
To find these points, we need to find the maximum and minimum value of the tangent function within the principal cycle.
The tangent function has a maximum value of +∞ and a minimum value of -∞. Therefore, there are no halfway points in this case.
In summary:
1) The interval for the principal cycle is (-π/3, π/3).
2) The equations of the vertical asymptotes are x = π/6 and x = π/2.
3) The coordinates of the center point are (0, 0).
4) There are no halfway points.
I hope this helps! Let me know if you have any more questions.