What is the domain and range of $f (x) = \sqrt{3 + x - 2}$ $f(x)= sqrt (3+x-2)$ ?

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Answer 1 · 2018-03-15T06:19:30

Domain: $x \geq (- 1)$ $x>= (-1)$ . Using interval notation: $[- 1, \infty)$ $[-1,oo)$

Range: $f (x) \geq 0$ $f(x)>= 0$ . Using interval notation: $[0, \infty)$ $[0,oo)$

Given:

$f (x) = \sqrt{3 + x - 2}$ $color(red)(f(x)=sqrt(3+x-2)$

Domain:

The domain of a function refers to a set of input values for which the function is real and defined.

$\sqrt{f (x)} = f (x) \geq 0$ $sqrt(f(x)) = f(x)>=0$

We solve $(3 + x - 2) \geq 0$ $(3+x-2)>=0$

We will keep the like terms together as a group.

$x - 2 + 3 \geq 0$ $x-2+3>=0$

$x + 1 \geq 0$ $x+1>=0$

Add $(- 1)$ $(-1)$ to both sides to simplify.

$x + 1 - 1 \geq 0 - 1$ $x+1-1>=0-1$

$\Rightarrow x \geq (- 1)$ $rArr x>=(-1)$

Hence, the domain is $x \geq (- 1)$ $x>=(-1)$ .

Using interval notation we can write the domain as $[- 1, \infty)$ $color(blue)([-1,oo)$

Range:

The range of a radical function is $f (x) \geq k, k = 0$ $f(x)>=k, k = 0$

Hence,

Range: $f (x) \geq 0$ $f(x)>=0$

Using interval notation the range can be written as $[0, \infty)$ $color(blue)([0, oo)$

Hope it helps.

What is the domain and range of f(x)= sqrt (3+x-2)f(x)=√3+x−2f(x)= sqrt (3+x-2)?