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

1 Answer
Oct 4, 2017

Answer:

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]}