#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)#
