What is the domain and range for #f(x)= x/(x^2-5x)#?

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Answer 1 · 2015-09-20T20:57:25

The domain of #f(x)# is #(-oo, 0) uu (0, 5) uu (5, oo)# and

the range of #f(x)# is #(-oo, -1/5) uu (-1/5, 0) uu (0, oo)#.

#f(x) = x/(x^2-5x) = x/(x(x-5)) = 1/(x-5)# with exclusion #x != 0#

The denominator of #f(x)# is zero when #x=0# or #x=5#.

Let #y = f(x) = 1/(x-5)#. Then #x = 1/y + 5#.

Therefore #y = 0# is an excluded value. Also #y = -1/5# is an excluded value, since it would result in #x = 0#, which is an excluded value.

So the domain of #f(x)# is #(-oo, 0) uu (0, 5) uu (5, oo)# and

the range of #f(x)# is #(-oo, -1/5) uu (-1/5, 0) uu (0, oo)#.