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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

#1**+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 ≠ 10^{7}

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

#1**+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 ≠ 10^{7}

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