If f(x)=x^2+5x+6 and g(x)=3x+1, what is the domain of f(x)/g(x)?
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`f(x)=x^2+5x+6`
`g(x)=3x+1`
`(f(x))/(g(x)) = (x^2+5x+6)/(3x+1)`
This function `(f(x))/(g(x))` will be undefinable when `(3x+1)=0` OR `x = -1/3` .
For every other x there is no problem with `(f(x))/(g(x))` .
So the domain of `(f(x))/(g(x))` is;
`x in R-[-1/3]`
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The domain of a function y = f(x) is all the values that x can take for which y = f(x) is real and defined.
For f(x)=x^2+5x+6 and g(x)=3x+1, the value of y = f(x)/g(x) = (x^2 + 5x +6)/(3x+1)
= (x^2 + 2x + 3x + 6)/(3x+1)
= (x(x + 2) + 3(x + 2))/(3x+1)
= ((x+3)(x+2))/(3x+1)
The value of the numerator is defined for all values of x, the expression (f(x))/(g(x)) is not defined if g(x) = 0
3x+1 = 0
x = -1/3
The domain of (f(x))/(g(x)) is the set of real values except x = -1/3
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