Let #h(x) = x/(x+4)# and #k(x)=2x-4#, how do you find (hºk)(x) and simplify?

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Answer 1 · 2015-09-06T16:36:34

#(h@k)(x) = (x-2)/(x)#

#(h@k)(x)# is the same as #h(k(x))#

In other words, simplify #h(2x-4)#.

To do this, we plug in #2x-4#, or replace every #x# in #h(x)# with #2x-4#.

#h(k(x)) = (2x-4)/((2x-4)+4)#

Then, we simplify:

#h(k(x)) = (2x-4)/((2x-4)+4)#

#h(k(x)) = (2x-4)/(2x-4+4)#

#h(k(x)) = (2x-4)/(2x)#

The numerator and denominator have a common factor of #2#, so we factor out #2# and cancel.

#h(k(x)) = (2x-4)/(2x)#

#h(k(x)) = (2(x-2))/(2(x))#

#h(k(x)) = (x-2)/(x)#

This is already simplified.

#(h@k)(x) = (x-2)/(x)#