pretest #24 and 25
I tried to do both problems and made very little progress. For 24 I started by graphing y=2^sec(x). The graph is periodic and contains vertical asymtotes. However, I wasn't sure how to interpret that into a domain and range. Any suggestions? Also, for 25 I really didn't know where to start. Can you say k=sin(x)? Is that even worthwhile?How do you incorporate the 1:5 ratio?
posted by Carrie Desmond at 12:13 PM
2 Comments:
for #24, as you said--the graph is periodic, which means x can be any real number, so the domain can be all reals. then for the range, since sec(x)=1/(cos(x)), and the range of the cos(x) graph is [-1,1], the range for sec(x) will be -1>=sec(x) and 1<=sec(x). for the range of 2^(sec(x)), first of all, it has to be greater than 0, no matter how small the value of sec(x) is, then when sec(x)<=-1, 2^(sec(x))<=1/2; when sec(x)>=1, 2^(sec(x))>=2. for conclusion, the range of 2^(sec(x)) is 1/2>=2^(sec(x))>0 and 2^(sec(x))>=2.
but when finding the domain of 2^(sec(x)) don't you have to take into consideration that you can't take secant when cosine = 0? so you cannot include (pi/2) and (3pi/2) in the domain, because for each of these values, cos(x)=0, and sec(x) is undefined.
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