If #s(x)=x^2+1# and #t(x)=x-3#, does #s(t(x))=t(s(x))#?

2 Answers
Alan P.
Mar 1, 2017

No

Explanation:

Given
#color(white)("XXX")color(red)(s(x))=color(red)(x^2+1)#
and
#color(white)("XXX")color(blue)(t(x))=color(blue)(x-3)#

Then
#color(white)("XXX")s(color(blue)(t(x)))=(color(blue)(t(x)))^2+1#
#color(white)("XXXXXXX")=(color(blue)(x-3))^2+1#
#color(white)("XXXXXXX")=x^2-6x+8#
and
#color(white)("XXX")t(color(red)(s(x)))=color(red)(x^2+1)-3#
#color(white)("XXXXXXX")=x^2-2#

Clearly #s(t(x))!=t(s(x))#

Kenneth B
Mar 1, 2017

No.
#s(t(x)) = x^2-6x+10#, while #t(s(x)) = x^2-2#

Explanation:

#s(t(x)) =(x-3)^2+1# (plug #t(x)# in the place of x in #s(x)#
#s(t(x)) = x^2-6x+9+1# (square #(x-3)#)
#s(t(x)) = x^2-6x+10# (simplify)

#t(s(x)) = (x^2+1)-3# (plug in #s(x)# in the place of x in #t(x)#)
#t(s(x)) = x^2-2# (simplify)

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