Question

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

Answer 1

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 `>=0`. so we start by solving the inequality:

`2x^2-5x+2>=0`

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

`(2x-1)(x-2) = 0=> x=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 `1/2` and to the right of the root `2`, we can find the domain directly i.e.

`-oo<=x<=1/2` and `2<= x <=oo`

So our domain is:

`( -oo , 1/2] uu [ 2 , oo)`

Minimum value of function is `0`

Maximum value:

`lim_(x->oo)(sqrt(2x^2 -5x+2)) ->oo`

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

`lim_(x->oo)(sqrt(2x^2))->oo`

So range is:

`[ 0 , oo)`

Graph of `sqrt(2x^2-5x+2)`:

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