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If f(x) =1/(logx) and g(x) = 1/(8+x), what is the domain of (fxg)(x)? 

 

a) {x e R | x > 0 and x can't equal -8} 

b) {x e R | x > 0 and x can't equal 1} 

c) {x e R | x > 0} 

d{ {x e R | x can't equal -9 and 1} 

Julius  Jan 18, 2018

Best Answer 

 #1
avatar+6943 
+2

(f × g)(x)  =  ( 1 / (log x) )( 1 / (8 + x) )

 

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

 

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

log x  ≠  0

x  ≠  100

x  ≠  1

 

Since  8 + x  is in a denominator, we have the restriction  8 + x  ≠  0

8 + x  ≠  0

x  ≠  -8               Notice that this is already accounted for with  x > 0

 

So the domain is all real  x  values such that  x > 0   and   x  ≠  1

hectictar  Jan 18, 2018
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1+0 Answers

 #1
avatar+6943 
+2
Best Answer

(f × g)(x)  =  ( 1 / (log x) )( 1 / (8 + x) )

 

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

 

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

log x  ≠  0

x  ≠  100

x  ≠  1

 

Since  8 + x  is in a denominator, we have the restriction  8 + x  ≠  0

8 + x  ≠  0

x  ≠  -8               Notice that this is already accounted for with  x > 0

 

So the domain is all real  x  values such that  x > 0   and   x  ≠  1

hectictar  Jan 18, 2018

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