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

Mar 15, 2018

Domain: $x \ge \left(- 1\right)$. Using interval notation:$\left[- 1 , \infty\right)$

Range: $f \left(x\right) \ge 0$. Using interval notation:$\left[0 , \infty\right)$

#### Explanation:

Given:

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.

$\sqrt{f \left(x\right)} = f \left(x\right) \ge 0$

We solve $\left(3 + x - 2\right) \ge 0$

We will keep the like terms together as a group.

$x - 2 + 3 \ge 0$

$x + 1 \ge 0$

Add $\left(- 1\right)$ to both sides to simplify.

$x + 1 - 1 \ge 0 - 1$

$\Rightarrow x \ge \left(- 1\right)$

Hence, the domain is $x \ge \left(- 1\right)$.

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

Range:

The range of a radical function is $f \left(x\right) \ge k , k = 0$

Hence,

Range: $f \left(x\right) \ge 0$

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

Hope it helps.