Given g(x) = Squareroot(x-1) find a)The domain of g; b)the range of g (inequalities like x> 2 should be typed as x>=2 etc..)
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if $\displaystyle g(x)=\sqrt{x-1}$ then the domain of g(x) is whenever $\displaystyle x-1\geq 0 $ which implies that the domain is $\displaystyle x \geq 1$. Similarly, for the range we know that $\displaystyle g(x) \geq 0 \quad \forall{x}$
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