let f(x)=(3x-7)/(x+1). Find the domain of f. Answer in interval form.
find the range of f. Answer in interval form
thanks!
f(x) = (3x - 7) / (x + 1)
The denominator cannot be zero, so we can't have x + 1 = 0 → x = -1
The domain of f is all values except -1 . (-∞ , -1) U (-1, ∞)
If you solve this for x , you will get x = (y - 7)/(y - 3) .
Now we can see that y cannot be 3 .
Also...the degree of the numerator = the degree of the denominator,
so there is a horizontal asymptote at y = 3/1 = 3 .
So the range of f is all values except 3 . (-∞ , 3) U (3, ∞)
f(x) = (3x - 7) / (x + 1)
The denominator cannot be zero, so we can't have x + 1 = 0 → x = -1
The domain of f is all values except -1 . (-∞ , -1) U (-1, ∞)
If you solve this for x , you will get x = (y - 7)/(y - 3) .
Now we can see that y cannot be 3 .
Also...the degree of the numerator = the degree of the denominator,
so there is a horizontal asymptote at y = 3/1 = 3 .
So the range of f is all values except 3 . (-∞ , 3) U (3, ∞)