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

Jun 4, 2017

Domain: $\left(- \infty , \infty\right)$
Range: $\left[- \frac{1}{4} , \infty\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{2} + x$

As with any polynomial, this is well defined for all values of $x$, so its domain is the whole of the Real numbers $\mathbb{R}$, in interval notation: $\left(- \infty , \infty\right)$

One way of finding the domain is to complete the square:

${x}^{2} + x = {\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4}$

Note that:

${\left(x + \frac{1}{2}\right)}^{2} \ge 0$

for any Real value of $x$, with equality when $x = - \frac{1}{2}$

So the minimum value of $f \left(x\right)$ is:

$f \left(- \frac{1}{2}\right) = {0}^{2} - \frac{1}{4} = - \frac{1}{4}$

Since $f \left(x\right)$ is continuous and unbounded, we can deduce that the range is $\left[- \frac{1}{4} , \infty\right)$

One way of proving that goes as follows.

Let:

$y = {x}^{2} + x = {\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4}$

Add $\frac{1}{4}$ to both sides to get:

$y + \frac{1}{4} = {\left(x + \frac{1}{2}\right)}^{2}$

Transpose and take the square root of both sides, allowing for both positive and negative square roots to get:

$x + \frac{1}{2} = \pm \sqrt{y + \frac{1}{4}}$

Subtract $\frac{1}{2}$ from both sides to find:

$x = - \frac{1}{2} \pm \sqrt{y + \frac{1}{4}}$

So, provided $y \in \left[- \frac{1}{4} , \infty\right)$, there is at least one value of $x$ such that $y = {x}^{2} + x$. That is: $y$ is in the range of $f \left(x\right)$.