What is the domain and range of f(x) = 4-x/3 and f(x)=+/- squareroot of 1+x/1-x

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embizze | High School Teacher | (Level 2) Educator Emeritus

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The domain of a function is the set of allowable inputs; the range is the set of possible outputs. The domain is generally all real numbers -- some restrictions include dividing by zero, taking an even root of a negative number, and taking a logarithm of a non-positive number.

(1) Given `f(x)=4-x/3` : There are no restrictions on the input x and no restrictions on the possible outputs (this is a straight line with slope -1/3) so:

The domain is all real numbers ( the set `(-oo,oo),"or"x in RR` ) and the range is all real numbers.

The graph:

** If you meant `f(x)=(4-x)/3` this is still a straight line with slope of -1/3; the y-intercept will be 4/3 instead of 4**

(2) Given `f(x)=+-sqrt((1+x)/(1-x))` : Note that this is not a function. Individually, `g(x)=sqrt((1+x)/(1-x))` and `h(x)=-sqrt((1+x)/(1-x))` are each functions.

There are two possible restrictions on the domain of the relation f(x): you cannot divide by zero and you cannot take a square root of a negative number in the reals.

Thus `x != 1` since you cannot divide by zero. Also we must have `(1+x)/(1-x)>=0` . A rational expression is positive if the numerator and denominator agree in sign -- both positive or both negative. `1+x>0` for `x > -1` and `1-x>0` for `x<1` so numerator and denominator are both positive on `-1<x<1` . Also `1+x<0=>x<-1` and `1-x<0=>x>1` so it is impossible for the denominator and numerator to both be zero.

Thus the domain is `-1<=x<1` .

As x approaches 1, the denominator approaches zero, so the rational expression approaches infinity (the numerator approaches 2, and 2 divided by an arbitrarily small number is arbitrarily large). If you take the negative of the square root, you will get a reflection across the x-axis. Thus the range is all real numbers.

The graph: