How do you find the domain and range for #y = sqrt(x^2 -3x +2)#?

1 Answer
Oct 22, 2017

Answer:

Domain: # x <= 1 and x >=2 or x| (-oo,1] uu [2,oo)#
Range: # y >=0 or y| [0,oo)#

Explanation:

#y= sqrt(x^2-3x+2) = sqrt((x-1)(x-2));# Domain: under

root should be #>=0 :. (x-1)(x-2)>=0#

When # 1 < x < 2 # sign of # y # is #(+)(-) = (-) :. < 0#

Therefore for #1 < x < 2 ; y # is undefined .

Domain: # x <= 1 and x >=2 or x| (-oo,1] uu [2,oo)#

Range: # y >=0 or y| [0,oo)# since square root of positive quantity

is also positive.

graph{(x^2-3x+2)^0.5 [-10, 10, -5, 5]}