What is the domain and range of #f(x)={ x^2 - 81 }/ {x^2 - 4x}#?

1 Answer
Dec 30, 2017

Answer:

#D_f=RR-{0,4}=(-oo,0)uu(0,4)uu(4,+oo)# , Range = #f(D_f)=(-oo,(81-9sqrt65)/8]uu[(81+9sqrt65)/8,+oo)#

Explanation:

#f(x)=(x^2-81)/(x^2-4x)#

In order for this function to be defined we need #x^2-4x!=0#

We have #x^2-4x=0# #<=># #x(x-4)=0# #<=># #(x=0,x=4)#

So #D_f=RR-{0,4}=(-oo,0)uu(0,4)uu(4,+oo)#

For #x##inD_f# ,

#f(x)=(x^2-81)/(x^2-4x)# #=# #((x-9)(x+9))/(x^2-4x)#

#f(x)=0 <=> (x=9,x=-9)#

  • #(x^2-81)/(x^2-4x)=y# #<=># #x^2-81=y(x^2-4x)#

#x^2-81=yx^2-4xy#

  • Adding #color(green)(4yx)# in both sides,

#x^2-81+4yx=yx^2#

  • Substracting #color(red)(yx^2)# from both sides

#x^2-81+4yx-yx^2=0# #<=>#

#x^2(1-y)+4xy-81=0#

This is quadratic equation for #x# so
#a=1-y#
#b=4y#
#c=-81#

We need #D=b^2-4*a*c>=0# #<=>#
#16y^2-4(1-y)*(-81)>=0# #<=>#
#16y^2+324(1-y)>=0# #<=>#
#16y^2-324y+324>=0# #<=>#
#4y^2-81y+81>=0#

#y_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

#=# #(81+-sqrt(6561-1296))/8#

#=# #(81+-sqrt(5265))/8#

#=# #(81+-9sqrt65)/8#

#4y^2-81y+81>=0# #<=># #(y<=(81-9sqrt65)/8# or #y>=(81+9sqrt65)/8)#

so, #f(x)<=(81-9sqrt65)/8# or #f(x)>=(81+9sqrt65)/8#

Which means, #f(D_f)=(-oo,(81-9sqrt65)/8]uu[(81+9sqrt65)/8,+oo)#