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

1 Answer
Oct 14, 2015

Answer:

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

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

Explanation:

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}#