What is the domain and range of #f (x) = (x^2-2)/(x^2-4)#?

2 Answers
Apr 19, 2017

Answer:

Domain and range of this function

Explanation:

Domain:
#x<-2# or #-2 < x < 2# or #x>2#

Range:
#f(x)<= 1/2# or #f(x)>1#

Apr 19, 2017

Answer:

#x inRR,x!=+-2#
#y inRR,y!=1#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be.

#"solve " x^2-4=0rArrx^2=4#

#rArrx=+-2larrcolor(red)" excluded values"#

#rArr"domain is " x inRR,x!=+-2#

To find the value that y cannot be find #lim_(xto+-oo)f(x)#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2-2/x^2)/(x^2/x^2-4/x^2)=(1-2/x^2)/(1-4/x^2)#

as #xto+-oo,f(x)to(1-0)/(1-0)#

#rArryto1larrcolor(red)" excluded value"#

#rArr"range is " y inRR,y!=1#

The graph of f(x) illustrates this.
graph{(x^2-2)/(x^2-4) [-10, 10, -5, 5]}