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

#1**+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 ≠ 10^{0}

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

#1**+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 ≠ 10^{0}

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