Answer: (-∞,-1) ∪ (0,+∞)
Step-by-step explanation: The representation fog(x) is a representation of composite function, meaning one depends on the other.
In this case, fog(x) means:
fog(x) = f(g(x))
fog(x) =
![3(x+(1)/(x) )-(1)/(x+(1)/(x) ) -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/tm345jzlhonbwkkwgtvedxzw9zpx1oazkn.png)
![fog(x)=3x+(3)/(x) -(1)/((x^(2)+x)/(x) ) -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/s5u6npb285xbpfjzvt8gfyaqjp8tpyrc9d.png)
![fog(x)=3x+(3)/(x) -(x)/(x^(2)+x) -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/w4b0ibezoqiqghhb5vaa008wbj2gl88tox.png)
![fog(x)=(3x^(2)(x^(2)+x)+3(x^(2)+x)-x-4x(x^(2)+x))/(x(x^(2)+x))](https://img.qammunity.org/2021/formulas/mathematics/high-school/uujvi00ddjyyc1ya022wjspo6cwko6xq6k.png)
![fog(x)=(3x^(4)+3x^(3)+3x^(2)+3x-x-4x^(3)+4x^(2))/(x(x^(2)+x))](https://img.qammunity.org/2021/formulas/mathematics/high-school/5eiivx1fjv9q2burs94ru97jdjqr47wwey.png)
![fog(x)=(3x^(4)-x^(3)-x^(2)+2x)/(x(x^(2)+x))](https://img.qammunity.org/2021/formulas/mathematics/high-school/xhn6mzik3x06hfw01e59qse5nrzirwkw26.png)
This is the function fog(x).
The domain of a function is all the values the independent variable can assume.
For fog(x), denominator can be zero, so:
![x(x^(2)+x) \\eq 0](https://img.qammunity.org/2021/formulas/mathematics/high-school/tvosx6xuzmhi392j2ttw74sy2l9z0ynolz.png)
If x = 0, the function doesn't exist.
![x^(2)+x \\eq0](https://img.qammunity.org/2021/formulas/mathematics/high-school/5all1c423x9rn716rko5pu5sq105af7iax.png)
![x(x+1) \\eq0](https://img.qammunity.org/2021/formulas/mathematics/high-school/3po33eb3ntvlijcgkm1np4idpr4zsiv5s4.png)
![x+1\\eq0](https://img.qammunity.org/2021/formulas/mathematics/high-school/s8s4njo80tqi8yedc4aaiunmjixs8j1i1c.png)
Therefore, the domain of this function is: -∞ < -1 or x > 0