# How do you find the domain and range of f(x)=sqrt(2x^2-5x+2)?

Oct 4, 2017

See below.

#### Explanation:

If we are going to stay in the realm of real numbers, then the expression inside the radical (This is called the radicand) must be $\ge 0$. so we start by solving the inequality:

$2 {x}^{2} - 5 x + 2 \ge 0$

Factoring to find roots. (This gives us a starting position ):

$\left(2 x - 1\right) \left(x - 2\right) = 0 \implies x = \frac{1}{2}$ or $x = 2$.

We can actually cheat a bit here. Notice the coefficient of ${x}^{2}$ is positive. This means it has a minimum value. The maximum value will then be above the $x$ axis and to the left of the root $\frac{1}{2}$ and to the right of the root $2$, we can find the domain directly i.e.

$- \infty \le x \le \frac{1}{2}$ and $2 \le x \le \infty$

So our domain is:

$\left(- \infty , \frac{1}{2}\right] \cup \left[2 , \infty\right)$

Minimum value of function is $0$

Maximum value:

${\lim}_{x \to \infty} \left(\sqrt{2 {x}^{2} - 5 x + 2}\right) \to \infty$

Since $2 {x}^{2}$ changes much more rapidly than $5 x$ we could have simplified the limit to:

${\lim}_{x \to \infty} \left(\sqrt{2 {x}^{2}}\right) \to \infty$

So range is:

$\left[0 , \infty\right)$

Graph of $\sqrt{2 {x}^{2} - 5 x + 2}$:

graph{sqrt(2x^2-5x+2) [-58.5, 58.57, -29.24, 29.23]}