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# combine functions

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42
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+-6

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

### 2+0 Answers

#1
-6
+2

x+4/x where x≠0, 2/3

The last answer is correct

May 13, 2019
#2
+8118
+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