Question 1209612
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f(5/x) is what we should focus on.
The input 5/x needs to be on the interval -1 < 5/x < 1, so it fits with the domain interval (-1,1).


You should find that -1 < 5/x < 1 leads to either x < -5 or x > 5.
But none of those values are in the interval -1 < x < 1.


Therefore the <font color=red>domain of g(x) is the empty set</font>.


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Some examples:


Let's try x = 0.5
g(x) = 5 - f(x) + f(5/x)
g(0.5) = 5 - f(0.5) + f(5/0.5)
g(0.5) = 5 - f(0.5) + <font color=red>f(10)</font>
g(0.5) = 5 - f(0.5) + <font color=red>undefined</font>
g(0.5) = <font color=red>undefined</font>
This shows that x = 0.5 is not in the domain of g(x).
f(10) is undefined since f(x) is only defined when -1 < x < 1.


Let's try x = 10
g(x) = 5 - f(x) + f(5/x)
g(10) = 5 - f(10) + f(5/10)
g(10) = 5 - <font color=red>f(10)</font> + f(0.5)
g(10) = 5 - <font color=red>undefined</font> + f(0.5)
g(10) = <font color=red>undefined</font>
We run into the same problem as before.
This shows that x = 10 is not in the domain of g(x).


I encourage the student to explore other examples.
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