Algebra 2 question

Question

What is the domain & range of 1/sqrt x+4

Answer 1

Emoni:

 

You must know how to write mathematical expressions carefully.  Here you must understand the difference between

 

1/sqrt x+4   interpreted as 1/[sqrt(x)] + 4]   OR interpreted as 1/[sqrt(x+4)]

 

The correct answer depends on what you mean by 1/sqrt x+4.

 

The crucial point is that you should choose the domain such that at no point in it the denominator of  the function is zero.

 

1) 1/sqrt x+4 means 1/[sqrt(x)] + 4] .

 

You know that for all x > 0, sqrt(X) > 0.  And so, for all x > 0, sqrt(x) + 4 > 0. 

 

So the domain here is {x : x > 0}

 

2) 1/sqrt x+4 means 1/[sqrt(x+4)].

You know that square-roots of negative numbers are not real numbers.  Further the denominator can not be zero. You achieve both if you take  x > - 4, because then x+4 > 0 and sqrt(x+4) is a positive real number.  So, for (2)

 

NOW the domain is {x: x > - 4}

 

The range in BOTH interpretations, is of the form 1 divided by a positive number, which is positive but can be very large. So the range is, in interval notation, (0, infinity) and in set notation,  {x: 0 < x < infintity)

 

Tell me Emoni, which interpretation of 1/sqrt x+4 you have/had in mind, choose the appropriate domain (either from (1) or from (2) and post a comment.  Next time write your mathematical expressions less ambiguously.

 

Dr. G.

 

 

Answer 2

f(x) = 1/√(x+4)

 

the denominator cannot be zero. x=-4 makes the denominator zero so x = -4 is not in the domain.

 

any number less than -4 makes the inside of the square root negative. We cannot take square root of a negative number so all numbers less than -4 are not in the domain. All numbers greater than -4 are acceptable.

 

domain x > -4

 

when x gets very large, the function approaches zero but never equals zero

 

when x is very close to -4, the function approaches infinity

 

range (0,∞)