How do you find #(h-g)(t)# given #h(t)=2t+1# and #g(t)=2t+2#?

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Answer 1 · 2017-05-26T13:56:39

#(h-g)(t)=-1#

#color(blue)("Preamble")#

#h(t)# is just an identifier for a particular process applied to the variable #t#

So #h(t)=2t+1 and h(s)=2s+1 and h(b)=2b+1#

So #g(t)=2t+2 and g(s)=2s+2 and g(b)=2b+2#
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#color(blue)("Answering the question")#

The question can also be written as: #h(t)-g(t)#

#h(t)->2t+1#
#g(t)->ul(2t+2)larr" subtract"#
#" "0t-1#

#(h-g)(t)=-1#