How do you find #(fog)(x)# given #f(x)=2x-5# and #g(x)=x+2#?

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Answer 1 · 2017-02-16T14:25:11

#(fog)(x)=2x-1#

Another way of looking at it:

set #g(x)->y_1=x+2...........Equation(1)#

set #f(x)->y_2=2x-5........Equation(2)#

#(fog)(x)# is such that as f is before the g in #(fog)(x)#

so wherever in #f(x)# you see #x# you substitute #y_1# That is; you are substituting #Equation(1)# into any #x" that is in "Equation(2)#

So: #(fog)(x)=color(green)(2(color(red)(y_1))-5)#

but #" "color(red)(y_1=x+2)#

So: #(fog)(x)=color(green)(2(color(red)(x+2))-5)#

#(fog)(x)=2x+4-5" "=" "2x-1#