How do you find the domain and range of #f(x) = x / (3x(x-1))#?

Question

Kennedy Adams · Accepted Answer

Domain: #RR#\#{0,1}#

Range: #RR#\#{-1/3,0}#

Domain: To find the domain of this rational function, simply look for the values of #x# that will make the the denominator #0#. #color(white)(XXX)3x(x-1)=0# #hArrcolor(white)(X)color(blue)(x=0)orcolor(red)(x=1)# Exclude 0 and 1 from the set of real numbers. The domain is #RR#\#{0,1}# Range: To find the range, start by isolating #x#. #color(white)(XX)y=x/[3x(x-1)]# #color(white)(XX)y=cancelx/[3cancelx(x-1)]# Note: When we cancel #x# here, we are adding a restriction that #x!=0#. The graph of this new equation has a horizontal asymptote #y=0#. Therefore, #0# is removed from the range. #color(white)(XX)y=1/[3(x-1)],x!=0# #color(white)(XX)y=1/[3x-3],x!=0# #color(white)(XX)y(3x-3)=1,x!=0# #color(white)(XX)3xy-3y=1,x!=0# #color(white)(XX)3xy=1+3y,x!=0# #color(white)(XX)x=(1+3y)/(3y),x!=0# Now take note that #x# must not be equal to #0# or #1#. We will substitute this into the equation to find out the values that #y# cannot be. #[1]color(white)(X)x=(1+3y)/(3y)# #color(white)([1]X)(0)=(1+3y)/(3y)# #color(white)([1]X)0=1+3y# #color(white)([1]X)3y=-1# #color(white)([1]X)y=-1/3# #[2]color(white)(X)x=(1+3y)/(3y)# #color(white)([2]X)(1)=(1+3y)/(3y)# #color(white)([2]X)cancel(3y)=1+cancel(3y)# #color(white)([2]X)0!=1# From [1], we know that #-1/3# is excluded from the range. [2] doesn't matter. And looking back at our note, #0# is also excluded from the range. Therefore, the range is #RR#\#{-1/3,0}#

Julia Adams · Answer

undefined Domain: $x \in \mathbb{R}, x \neq 0, x \neq 1$

Range: $f(x) \in \mathbb{R}, f(x) \neq 0$