What are the domain and range of $y = 1 + \sqrt{x + 3}$ $y= 1+sqrt(x+3)$ ?

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Answer 1 · 2018-01-13T03:36:11

Domain: $[- 3, + \infty)$ $[-3, +oo)$ Range: $[1, + \infty)$ $[1,+oo)$

$y = 1 + \sqrt{x + 3}$ $y= 1+sqrt(x+3)$

$y$ $y$ is defined where $x + 3 \geq 0 \to x \geq - 3$ $x+3>=0 -> x>=-3$

Hence, the domain of $y$ $y$ is $[- 3, + \infty)$ $[-3,+oo)$

$y$ $y$ has a minimum value of 1 where $x = - 3$ $x=-3$

$y$ $y$ has no finite upper bound.

Hence, the range of $y$ $y$ is $[1, + \infty)$ $[1,+oo)$

We can deduce these results from the graph of $y$ $y$ below.

graph{1+sqrt(x+3) [-10, 10, -5, 5]}

What are the domain and range of y=1+√x+3y= 1+sqrt(x+3)?