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

1 Answer
Somebody N.
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 >=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]}

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