Find the domain and range of : f(x)= 1/ x+1?

Answer 1

I presume you mean
f(x) = 1/(x+1)

Another way of stating domain is, "What values can x be?"
In this case, the restriction lies in the fact that you cannot divide by 0. Therefore, all you have to do is equate the denominator to 0 and see what you get; the answer you get is what x CAN'T be, leaving the rest.

x+1 = 0 implies x = -1

Therefore, your domain is (-infinity, -1) U (-1, infinity)

To get the range, you have to ask yourself; if y = 1/(x+1), what can y be? The answer to that is that it can't be 0.

Therefore, your range is (-infinity, 0) U (0, infinity)

Consider a function in this form:

f(x) = A/(x-h) + K

The K will be your horizontal asymptote, and the range will end up being (-infinity, K) U (K, infinity). In the case above, there's actually a 0 there.

f(x) = 1/(x+1) + 0

and that's why our range is the way it is.

Answer 2

I cannot tell if the function is 1/(x+1) or (1/x) + 1. I will do both.

If the function is 1 / ( x + 1 ), we must be concerned with the x's that make the denominator zero, since that is very bad. Thus, an x-value of -1 is bad. So, the domain is {all real x except x = -1}. The range is the possible values the function can take. By plugging in all possible values for x, we see that the function can take on any value except 0. This makes sense because a fraction can only equal 0 if the numerator equals 0. This numerator is always 1, excluding 0 as a possible answer. Thus, the range is {all real y except y = 0}.

If the function is ( 1 / x ) + 1, then we can rewrite the function as
( x + 1 ) / x. We must still be concerned with values of x that make the denominator 0. This only happens if x = 0. Thus, the domain here is {all real x except x = 0}. The range is all possible values that the function can take. For the same reason as above, (1 / x) can never equal 0. This implies that ( 1 / x ) + 1 can never eqaul 1. In fact, this is the only value that the function cannot achieve. Thus, the range of this function is {all real y except y = 1}.

DONE.

Answer 3

Use limits to find values as x approaches zero from the left and right side of the function:
LIM x-0
As x increases, the function appears to reach infinity until the denominator is undefined at infinity; as x decreases, there is also discontinuity at x=0, making this point undefined. When x is less than 0, the same thing happens on the other side when negative infinity is approached.
(X> -infinity, not=0, X< +infinity)
(F(x) > (1-1/(all x values > -infinity), not = 0, (1-1/(all x values < + infinity)

Answer 4

the realm is [0, 4/3], thinking the very shown truth that we've a function f(x) defined only in [0, 4], so g(x) is defined on condition that f(3x) is defined. for this reason you be sure 0 <= 3x <= 4, which elements the above area. the shape is [a million, 3] and does no longer change thinking the very shown truth that we even with the fact which have f(x) over an identical era.

Answer 5

Domanin (AKA D:) is all the numbers that x can equal. x cannot equal -1 because -1+1=0 which makes the question divided by zero which is illegal. Range (AKA R:) is all the numbers that can equal y (AKA f(x)) which is all real numbers. If you have to put it in a different form then its:

D: (negative infinity, -1)U(-1, infinity)
R: (negative infinity, infinity)

Answer 6

The previous poster assumed that you meant 1/(x+1), is that right?

Anyway, f(x) is never zero.

lim x -> -inf f(x) -> 0- , ie, is just below the x axis

lim x -> +inf f(x) -> 0+, ie, is just above the x axis

That means that the x axis is an asymptote.

Answer 7

D=R-{0} \R=R(it never collids with y=infinite and y=1 but the limit is equal)

Answer 8

domain is all real numbers except negative one
range is all real numbers I believe