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

The domain of a function represents the possible input values, i.e. values of `x`, for which the function is 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 `0` is undefined. This implies that any value of `x` that will make

`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)))) ->` the quadratic 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 `0` and the function is undefined. For any other value of `x`, `f(x)` will be defined.

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 `x = -9` and `x = 4/3`, i.e. the function ahs two 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)`