The domain of a function y = f(x) is the set of values x can take for which y is real and defined.

For the function f(x)=(5x+4)/(x^2+3x+2), the numerator is defined for all values of x. The function is indeterminate when the denominator x^2 + 3x + 2 is equal to 0. The values of x where this is the case have to be eliminated.

x^2 + 3x + 2 = 0

x^2 + 2x + x + 2 = 0

x(x + 2) + 1(x + 2) = 0

(x + 1)(x + 2) = 0

x = -1 and x = -2

At the values x = -1 and x = -2 the function y = f(x) is not defined.

This gives a domain of R - {-1, -2}

To find the domain of f(x) = (5x+4)/(x^2+3x+2).

Solution :

We shall find the values of x for which the denominator x^2+3x+2 becomes zero. And for that x the function becomes undefined with a jump from -infinity to positive ifinity.

x^2+3x+2 = (x+2)(x+1) becomes zero when x+2 = 0 or x+1 = 0. Or the denominator x^2+3x+2 bemes zero for x = -2 or x =-1.

So x wont take values -2 and -1. And x tacan take any other real values.

So

We'll establish the domain knowing the fact that the division by 0 is not allowed.

For this reason, we'll find out first, the x values for the denominator is cancelling.

To find out these values, we'll have to calculate the roots of the quadratic equation from the denominator.

x^2+3x+2=0

x1=[-3+sqrt(9-8)]/2

x1=(-3+1)/2

x1=-1

x2=(-3-1)/2

x2=-2

From here we conclude that the function is not defined for x=-1 and x=-2.

**So, the domain of definition is: R - {-2;-1}**