Question

Consider the following function f(x)=2x/3x^2-3 what is the domain of the function and what is the range of the function

Answer 1

Answer:

Domain: $x\ne \pm1$

Range: $y\in R$

Step-by-step explanation:

The given function is $f(x)=\frac{2x}{3x^2-3}$

This rational function is defined for all real values except where

$3x^2-3=0$

$x^2-1=0$

$x^2=1$

$x=\pm \sqrt{1}$

$x=\pm 1$

The domain is all real numbers except $x=\pm 1$

The range is all real numbers.

Answer 2

Answer:

domain (-infinity, -1) U(-1,1) U (1, infinity)

range is (-infinity, infinity)

Step-by-step explanation:

Domain is the set of all x  values that makes the function defined

We need to look at x  values that makes the function undefined

When denominator =0 then function is undefined

we look at the x values that makes the denominator 0. Set the denominator =0  and solve for x

3x^2 - 3=0

3x^2 = 3

divide by 3 on both sides

x^2 = 1

take square root on both sides

x= +-1 , so x=+1  and x=-1

x=-1  and +1 makes the denominator 0. so we ignore -1  and +1 in the domain

Domain is set of all numbers except -1 and +1

(-infinity, -1) U(-1,1) U (1, infinity)

Range is the set of y values

for every value of x  we get an y value

we have only 2x in the numerator . so there is no restriction for y values

So y is all real numbers

range is (-infinity, infinity)