Question

find the domain of f( x )= 1/1-2cosc​

Answer 1

Answer:

5

the domain of ln(z) is z>0. So we need 1−2cos(x)>0. so

−2cos(x)>−1

2cos(x)<1

cos(x)<12

if we graph cos(x) we see that cos(x)≥12 for x∈[−π3,π3] but we need to exclude this to keep cos(x)<12, so we take these x values out of this domain. But this region occurs periodically so we need to exclude x∈[2nπ−π3,2nπ+π3] where n∈Z

so the domain is

x∈R∖[2nπ−π3,2nπ+π3]

As for the range: the max value of f(x) occurs when cos(x) is a minimum, so the minimum value of cos(x) in the domain is cos(x)=−1, so putting this into f(x), we get ln(1−2(−1)) = ln(3) As for the lower bound of the range, cos(x) never actually reaches 12 but it does get infintiely closer to 12 so 1−2cos(x) gets closer to 0 and from our knowledge of logarithms and exponentials, exp(z) gets closer to 0 as z gets closer to -infinity, so f(x) will tend to -infinity, therefore the range is (−inf,ln(3)]

Step-by-step explanation: