Find the domain f=(5x+4) / (x^2+3x+2)

Expert Answers
hala718 eNotes educator| Certified Educator

f(x)= (5x+4)/(x^2+3x+2)

First we need to determine x values where f(x) is not defined.

f is not defined when x^2+3x+2 =0 (since f is a ratio)



x= {-1,-2}

Then f domain if R-{-1,-2}

tonys538 | Student

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}

neela | Student

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.


giorgiana1976 | Student

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.







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}