Given f(x)=3x2+10x−8 and g(x)=3x2−2x .
What is (f/g)(x) ?
−5x+4/x where x≠0, 2/3
x/−5x+4 where x≠0, 4/5
x/x+4 where x≠0, 4
x+4/x where x≠0, 2/3
Given \(f(x)=3x^2+10x-8\) and \(g(x)=3x^2-2x\)
What is \((\frac{f}{g})(x)\) ?
--------------------------
\((\frac{f}{g})(x)\,=\,\frac{f(x)}{g(x)}\)
Substitute \(3x^2+10x-8\) in for \(f(x)\) and substitute \(3x^2-2x\) in for \(g(x)\)
\((\frac{f}{g})(x)\,=\,\frac{3x^2+10x-8}{3x^2-2x}\)
Split 10x into two terms such that their coefficients add to 10 and multiply to -24
\((\frac{f}{g})(x)\,=\,\frac{3x^2+12x-2x-8}{3x^2-2x}\)
Factor the numerator and denominator.
\((\frac{f}{g})(x)\,=\,\frac{3x(x+4)-2(x+4)}{3x^2-2x}\\~\\ (\frac{f}{g})(x)\,=\,\frac{(x+4)(3x-2)}{x(3x-2)} \)
Cancel the common factor of (3x -2) and note that 3x - 2 ≠ 0 , that is, x ≠ 2/3
\( (\frac{f}{g})(x)\,=\,\frac{x+4}{x}\qquad\text{and}\qquad x\neq\frac23\)
Since x is in the denominator we can also note that x ≠ 0