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!

$0 <= 2x <= 1/2$

$0 <= x <= 1/4$

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