What is the range and domain of `f(x) = (3x^2)/(sqrt(x+1)*(x-1))`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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For a function f(x), the domain is the set of values that x can take on for which f(x) is defined and real. The range is all the values of f(x) for x lying in the domain.

The function of which the range and domain has to be determined is:

`f(x) = (3x^2)/(sqrt(x+1)*(x-1))`

For f(x) to be real and defined, `x + 1 > 0` as the square root of a negative number is complex and the denominator cannot be equal to 0. Also, `x - 1 ! = 0`

`x + 1 > 0`

=> `x > -1`

`x - 1!=0`

=> `x != 1`

This gives the set of values that x can lie in as `(-1, oo)-{1}`

For -1< x < 1, f(x) is negative and for x > 1, f(x) is positive. The range of the function is the set of real numbers R.

The function `f(x) = (3x^2)/(sqrt(x+1)*(x-1))` has a domain `(-1, oo)-{1}` and the range is R