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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

 May 13, 2019
edited by BellaNight7310  May 13, 2019
 #1
avatar-6 
+2

 

  x+4/x where x≠0, 2/3

The last answer is correct

 May 13, 2019
 #2
avatar+8852 
+3

Given    \(f(x)=3x^2+10x-8\)    and    \(g(x)=3x^2-2x\)

What is \((\frac{f}{g})(x)\)   ?

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\((\frac{f}{g})(x)\,=\,\frac{f(x)}{g(x)}\)

                                              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

 May 14, 2019

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