Find the domain f=(5x+4) / (x^2+3x+2)
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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^2+3x+2=0
(x+1)(x+2)=0
x= {-1,-2}
Then f domain if R-{-1,-2}
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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}
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}
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