# Thread: Find the domain and range

1. ## Find the domain and range

g(x) = sqrt.(25 - x^2)
r(x) = |2x - 3|

plz help me thanks!

2. Hello, NeedHelp18!

Find the domain and range.

$\displaystyle 1)\;\;g(x) \:= \:\sqrt{25 - x^2}$

We can see that $\displaystyle (25-x^2)$ must not be negative.

We have: .$\displaystyle 25-x^2 \:\geq \:0$

Then we have: .$\displaystyle -x^2 \:\geq \:-25$

Multiply by -1: .$\displaystyle x^2 \:\leq \:25$

And we have: .$\displaystyle |x| \:\leq \:5$

. . $\displaystyle \boxed{\text{Domain: }\;-5 \:\leq\:x\:\leq\:5}$

We see that $\displaystyle g(x)$ is never negative: .$\displaystyle g(x) \:\geq\:0$

When $\displaystyle x=0,\;g(0) \:=\:\sqrt{25} \:=\:5$ (maximum value)

. . $\displaystyle \boxed{\text{Range: }\;0 \:\leq \:g(x) \:\leq \:5}$

$\displaystyle 2)\;\;r(x) \:= \:|2x - 3|$

We can use any value for $\displaystyle x.$

. . $\displaystyle \boxed{\text{Domain: }\:(-\infty,\:\infty)}$

An absolute value is never negative: .$\displaystyle r(x) \:\geq\:0$

As $\displaystyle x \to \pm\infty,\;r(x) \to \infty$

. . $\displaystyle \boxed{\text{Range: }\;r(x) \:\geq 0}$