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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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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's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}

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