Will you please find domain and range of two functions: f(x) = 3x /(x + 2) and
f (x) = -2x / (x-3) , which is the inverse of the first function.
so find the domain of both functions and you should have everything.
the domain is the set of all x-values for which the function is defined. so find what x's we CANNOT have, and then the domain is all x's but those. (Hint: what can we not do with a fraction?)
note, the range is the set of y-values for which the function is defined. so when stating the range, make sure to do it in terms of y
Thank you for taking time and explaining all of this. I have solved the problem before I posted a question, but my answer does not coincide with the answer in the textbook.
f(x) = 3x /(x + 2) Domain: All real numbers except x = -2 ; Range: All real numbers except y = 3 ( my reasoning: y = 3 is a horizontal asymptote and the function does not cross the horizontal asymptote, because when I tried to solve 3x /(x + 2) = 3 , I got 3x = 3x + 6, which does not make sense
f(x) = -2x / (x-3) Domain: All real numbers except x = 3 ; Range: All real numbers except y = -2 as y = -2 is a horizontal asymptote of f(x) = -2x / (x-3) and does not cross the function.
Is this correct? The textbook has a different answer for the range, I do not have it with me right now, I will post it later, but will you please check whether my answers are correct.
it seems to me that you just tried to pick the horizontal asymptotes out of the blue. you could have found them directly using limits. but good stuff
How could I pick up horizontal asymptotes out of the blue correctly? To find the horizontal asymptotes for the rational functions with the same power of x in the nominator and the denominator, I used the coefficients of x in the nominator and the denominator, so for the function f(x) = 3x /(x + 2) the horizontal asymptote is y = 3/1, and for the function f(x) = -2x / (x-3) the horizontal asymptote is y = –2/1.
Does it make any sense?
Thank you again.