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how to find domain of real-valued function ( g o h) g(x) 4x-1 h(x) sq rt (x-4) close sq rt
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interval notation
is the question h(x)=sqrt(x-4) ? or some thing more with h(x)?
\[(g\circ h)(x) = g(h(x)) = 4(\sqrt{x-4}) - 1\]
You just plug in h(x) for x in g(x)
i got that much but what is the domain ??
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so here for the square root to be defined x-4>=0 and hence x>=4 therefore the domain is x>=4
so in interval notation (4, \[\infty\]
whoops lol (4, infinity]
?
\[x \in [4,\infty)\]
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x can be 4, and any other number up to (but never including) infinity.
thnks
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