# Find the largest domain for the following function: g(x) = (x^2 - 10)/3x^2 - 12what is the largest domain of this function

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In this case, the denominator of the function impose the domain of definition.

We'll impose the constraint of existence of the function:

3x^2 - 12 different from 0

In other words, the domain of definition will contain x values that don't cancel the denominator.

We'll compute the roots of the equation

3x^2 - 12 = 0

We'll divide by 3 both sides:

x^2 - 4 = 0

Since it is a difference of squares, we'll write it using the formula:

a^2 - b^2 = (a-b)(a+b)

x^2 - 4 = (x-2)(x+2)

But x^2 - 4 = 0, so

(x-2)(x+2) = 0

We'll put each factor as 0:

x - 2 = 0

x = 2

x + 2 = 0

x = -2

**The domain of definition is the real set R, rejecting the roots of the equation from denominator.**

**R - {-2 ; +2}**

To find the largest domain of g(x) = (x^2-10)/(3x^2-12)

The largest domain should have all the set of values of x for which the function is defined and the values of the expression is real.

g(x) = (x^2-10)/((3x^2-12).

The denominator is 3x^2-12. The denominator 3x^2-12 becomes 0 when 3x^2-12 = 0. Or

3x^2 = 12. Or

x&2 = 12/3 = 4.

x^2= 4.

We take the sqre root.

x1 = sqrt4 . Or x = 2

x2 = -sqrt4. Or x2 = -2.

So when x = 2 or x = -2, the denominator becomes zero. Therefore the function g(x) becomes undefined like (2^2-10)/0 at x= 2 or g(x) = (-2)^2-10)/ 0 at x = -2.

Therefore x takes all values defined by x <-2 or -2< x< 2 or x >2. This , we denote by { R } - {-2) - {2}. This implies that except x = -2 and x = 2 , x takes all other real values.

The domain of a function is defined as the values of x for which the function takes real values.

Here the function we have is g(x) = (x^2 - 10)/ (3x^2 - 12). [I am assuming that the denominator is (3x^2 -12)

Simplifying the function:

g(x) = (x^2 - 10)/ (3x^2 - 12)

=> g(x) = (x^2 - 10)/ 3(x^2 - 4)

using the relation that x^2 - a^2 = (x-a)(x+a)

=> g(x) = (x^2 - 10)/ 3(x - 2) (x+2)

Now we see that g(x) has real values for all values of x except x= 2 and x= -2 . At x = 2 and x= -2 the value of g(x) is not defined.

**So the domain of this function is all real values of x except x =2 and x=-2.**