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If f(x) = log x and g(x) = 1/(x-7), what is the domain of g(f(x))? 

 

So g(f(x)) = 1/(logx-7) 

 

I was a bit confused with the domain of a reciprocal function... 

 

Options: 

a) {x e R| x > 0} 

b) {x e R | x can't equal 7} 

c) {x e R | x > 0 and x can't equal 10 000 000} 

d) {x e R | x can't equal 10 000 000} 

Julius  Jan 18, 2018

Best Answer 

 #1
avatar+6250 
+2

g( f(x) )   =   1 / ( log x - 7 )

 

There are two things that we need to consider.

 

Since  log x  is part of the function, we have the restriction  x > 0 .

 

Since  log x - 7  is in a denominator, we have the restriction  log x - 7 ≠ 0 .

log x  ≠  7

x  ≠  107

x  ≠  10 000 000

 

So the domain is all real  x  values such that  x > 0  and  x  ≠  10 000 000

hectictar  Jan 18, 2018
Sort: 

1+0 Answers

 #1
avatar+6250 
+2
Best Answer

g( f(x) )   =   1 / ( log x - 7 )

 

There are two things that we need to consider.

 

Since  log x  is part of the function, we have the restriction  x > 0 .

 

Since  log x - 7  is in a denominator, we have the restriction  log x - 7 ≠ 0 .

log x  ≠  7

x  ≠  107

x  ≠  10 000 000

 

So the domain is all real  x  values such that  x > 0  and  x  ≠  10 000 000

hectictar  Jan 18, 2018

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