Respuesta :
Answer:
Domain: All values (-∞,+∞) except x = [tex]\left(\frac{\pi }{3}n,\:\frac{\pi }{3}+\frac{\pi }{3}n\right) [/tex] for all real integers.
Range: (-inf, inf)
Period: pi/3
Vertical Asymptotes: n*pi/3 for all real values of n
Step-by-step explanation:
Domain
- For domain we equate the denominator of the equivalent fraction to zero and solve for x.
- Denominator: sin (3x) = 0
- Solve for x
- x = (…, -2*pi/3, - pi/3, 0, pi/3, 2*pi/3,…)
- Hence we can condense the domain based on a pattern as follows:
x = are all values from - infinity to + infinity except x =[tex]\left(\frac{\pi }{3}n,\:\frac{\pi }{3}+\frac{\pi }{3}n\right) [/tex] for all values of n
Range
- What are the values the function outputs when its defined as per domain above.
- input the end-points of the domain in f(x) evaluated above: [tex]\left(\frac{\pi }{3}n,\:\frac{\pi }{3}+\frac{\pi }{3}n\right) [/tex] for all values of n.
- Now solve for x = 0 , y = ? and x = pi/3, y = ?
- We see the denominator approaches 0 at end points hence entire function approaches -∞ and ∞ !
Hence, Range = (-∞,∞)
Period
- The period of a cotangent function can be evaluated by a formula as follows:
- f(X) = a cot (bx)
- Period = Pi / modulus (b)
- Hence, from the given function we can evaluate accordingly as follows:
Period = Pi / 3
Vertical Asymptotes:
- Vertical Asymptotes exist for rational fractions and logarithms only.
- For values of x for which a fraction is undefined are denoted as vertical asymptotes as we scrutinize on the denominator.
- In our case, values of x when sin (3x) = 0
- From the part of domain evaluated above we got pi*n/3 for all values of n.
Hence, our two vertical asymptotes can be x = 0 and x = pi/3 with n = 0 and n = 1, respectively. Other possibilities can be evaluated by using other real values for n.
