Consider the function $z = \ln{(y + 1)}+\sqrt{x-3}$. Find the domain and range, and sketch the domain in the plane.
2026-03-31 11:11:09.1774955469
Domain and Range problem(plane)
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See that $\sqrt{a}$ is defined on the real plane for all $a\geq0$. So, $\sqrt{x-3}$ exists only for $\boxed{x\geq3}$. Then, $ln(b)$ is defined for all $b>0$. So, $\boxed{y>(-1)}$. The sketching and the range parts are trivial now that the domain is given.