Find the domain and range

**NeedHelp18** · October 14th 2008, 02:54 PM

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

plz help me thanks!

**Soroban** · October 14th 2008, 03:49 PM

Hello, NeedHelp18!

Find the domain and range.

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

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

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

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

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

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

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

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

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

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

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

We can use any value for $x.$

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

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

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

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

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