How do you find the zeroes for #R( x) = (3x)/(x^2 - 2x)#?

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Answer 1 · 2015-06-05T23:58:23

#R(x)=(3x)/(x^2-2x) = (3cancel(x))/((x-2)cancel(x)) = 3/(x-2)#

with restrictions #x!=0# and #x!= 2#.

When #x < 0# or #0 < x < 2# we have #x-2 < 0#

so #3/(x-2) < 0#

In fact #3/(x-2) -> 0_-# as #x -> -oo#

When #x > 2# we have #x-2 > 0#

so #3/(x-2) > 0#

In fact #3/(x-2) -> 0_+# as #x -> oo#

So #R(x) != 0# for all #x in RR#

graph{3x/(x^2-2x) [-10, 10, -5, 5]}