# What is the domain of #g(x) = (x+5)/(3x^2+23x-36)# in set notation?

##### 1 Answer

#### Answer:

#### Explanation:

The **domain** of a function represents the possible input values, i.e. values of *defined*.

Notice that your function is actually a fraction that has two rational expressions as its numerator and denominator, respectively.

As you know, a fraction that has a denominator equal to **undefined**. This implies that any value of

#3x^2 + 23x - 36 = 0#

will **not** be part of the domain of the function. This quadratic equation can be solved by using the **quadratic formula**, which for a generic quadratic equation

#color(blue)(ul(color(black)(ax^2 + bx + c = 0)))#

looks like this

#color(blue)(ul(color(black)(x_(1,2) = (-b + -sqrt(b^2 - 4 * a * c))/(2 * a)))) -># thequadratic formula

In your case, you have

#{(a = 3), (b = 23), (c = -36) :}#

Plug in your values to find

#x_(1,2) = (-23 +- sqrt( 23^2 + 4 * 3 * (-36)))/(2 * 3)#

#x_(1,2) = (-23 +- sqrt(961))/6#

#x_(1,2) = (-23 +- 31)/ 6 implies {(x_1 = (-23 - 31)/6 = -9), (x_2 = (-23 + 31)/6 = 4/3) :}#

So, you know that when

#x = -9" "# or#" " x = 4/3#

the denominator is equal to *undefined*. For **any other value** of

This means that the domain of the function in *set notation* will be

#{ x in RR | x < -9 or -9 < x < 4/3 or x > 4/3}#

graph{(x+5)/(3x^2 + 23x - 36) [-14.24, 14.23, -7.12, 7.12]}

As you can see from the graph, the function is not defined for *vertical asymptotes* in those two points.

Alternatively, you can write the domain as

#x in RR "\" {-9, 4/3}#

In *interval notation*, the domain would look like this

#x in (-oo, - 9) uu (-9, 4/3) uu (4/3, + oo)#