What is the domain of #fog(x)# given #f(x)=sqrt(x-2)# and #g(x)=1/(2x)#?

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Answer 1 · 2017-02-14T14:19:45

Dom #h =(0, 1/4]#

Fog is a MTG card.

#f \circ g# applied in #x#, is equal to #f(g(x))#.

#f(1/(2x)) = sqrt{1/(2x) - 2} = h(x)#

So, we can't divide by #x = 0#, and we can't take sqrt of negatives.

#1/(2x) - 2 >= 0#

#1/(2x) >= 2#

You see the hyperbola!

#0 <= 2x <= 1/2#

#0 <= x <= 1/4#