# How do you find the domain and range of f(x)=x^2+2x+2?

Nov 13, 2017

Domain: $x \in \mathbb{R}$
Range: $f \left(x\right) \in \left[1 , + \infty\right) , \mathbb{R}$

#### Explanation:

$f \left(x\right) = {x}^{2} + 2 x + 2$ is defined for all Real values of $x$
$\Rightarrow$ Domain is $\mathbb{R}$

$f \left(x\right) = {x}^{2} + 2 x + 2$ has a minimum value of $f \left(x\right) = 1$ but no maximum value
$\Rightarrow$ Range is $\left[1 , + \infty\right) , \mathbb{R}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

How do we know that $f \left(x\right)$ has a minimum value of $1$?

Since $f \left(x\right)$ is a quadratic with a coefficient of $x$ which is greater than zero,
we know that it has a minimum.
We can find this minimum by taking the derivative of $f \left(x\right)$, setting that to zero, solving for $x$, and then using that value to find $f \left(x\right)$ at that value:
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 2 x + 2 = 0$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = - 1$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow f \left(x = - 1\right) = {\left(- 1\right)}^{2} + 2 \cdot \left(- 1\right) + 2 = 1$

If it helps here is a graph of $f \left(x\right) = {x}^{2} + 2 x + 2$ to help verify this:
graph{x^2+2x+2 [-3.61, 2.55, -0.512, 2.568]}