How do you find the domain and range for $y = {(x + 3)}^{0.5}$ $y= (x+3)^0.5$ ?

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Answer 1 · 2018-05-29T17:06:55

Domain: ${x ∣ x \geq - 3}$ ${x|x>=-3}$ or $[- 3, \infty)$ $[-3, oo)$

Range: ${y ∣ y \geq 0}$ ${y|y>=0}$ or $[0, \infty)$ $[0, oo)$

$y = {(x + 3)}^{0.5}$ $y= (x+3)^0.5$

$y = {(x + 3)}^{\frac{1}{2}}$ $y= (x+3)^(1/2)$

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

So the domain will be all numbers where the terms under the radical are not negative (otherwise the solution is imaginary).

$x + 3 \geq 0$ $x+3>=0$

$x \geq - 3$ $x>=-3$

Domain: ${x ∣ x \geq - 3}$ ${x|x>=-3}$ or $[- 3, \infty)$ $[-3, oo)$

Now the range at $x = - 3; y = 0$ $x=-3; y=0$ but it will always be greater than or equal to 0.

Range: ${y ∣ y \geq 0}$ ${y|y>=0}$ or $[0, \infty)$ $[0, oo)$

Here is the graph:

graph{sqrt(x+3) [-6, 14, -1.28, 8.72]}

How do you find the domain and range for y= (x+3)^0.5y=(x+3)0.5y= (x+3)^0.5?