Not sure about this question...
1. State the range for the function f(x) = -log 1/2(base) (x-1) + 2
..help?
6 answers
I think its (-infinity, infinity)
You know that the range for log(x) is all real numbers.
Similarly, the range for log_2(x) is all real numbers, since log_2(x) = lnx/ln2
log_(1/2)(x) = -log_2(x) since 1/2 = 2^(-1)
So, log_(1/2)x = -lnx/ln2
log_(1/2)(x-1) is just the same graph, shifted right by 1
log_(1/2)(x-1)+2 is the same graph, shifted up by 2
The range is still all reals.
Visit wolfamalpha.com and type in
log_2(x-1)+2
Similarly, the range for log_2(x) is all real numbers, since log_2(x) = lnx/ln2
log_(1/2)(x) = -log_2(x) since 1/2 = 2^(-1)
So, log_(1/2)x = -lnx/ln2
log_(1/2)(x-1) is just the same graph, shifted right by 1
log_(1/2)(x-1)+2 is the same graph, shifted up by 2
The range is still all reals.
Visit wolfamalpha.com and type in
log_2(x-1)+2
Got it!
The range is basically all the values that can come out from plugging something in (in other words, every answer that the equation can have). In order to find the range, you need to graph this equation. Once you have the graph, find which point the asymptote(the dashed line) lies. This is the first value that goes into the range. Next, you find which side the actual line goes to. If it goes to the right, then the range is (y coordinate of asymptote, ∞). If it goes to the left, then it is (-∞, y coordinate of asymptote) hope it helps :)
ps. i like your username xD
ps. i like your username xD
well nevermind ._.
Thanks for the compliment haha
Been using this username on Jiskha for 5 years now
Been using this username on Jiskha for 5 years now
1 answer