How do you find the domain and range of #h(x)= 10/(x^2-2x)#?

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Answer 1 · 2015-07-22T10:14:39

Analyse behaviour of the denominator and hence #h(x)# to find:

domain #= (-oo, 0) uu (0, 2) uu (2, oo)#

and

range #= (-oo, -10] uu (0, oo)#

#h(x) = 10/(x^2-2x) = 10/(x(x-2)) = 10/((x-1)^2-1)#

The denominator is zero when #x=0# or #x=2#, so #h(x)# is undefined for those values of #x#.

When #x in (0, 2)#, #x^2-2x in [-1,0)#,

so #h(x) in (-oo, -10]#

When #x in (-oo, 0) uu (2, oo)#, #x^2-2x in (0, oo)#,

so #h(x) in (0, oo)#