Composing functions just means to feed the ouput of the inner one as the input for the outer one.
So, the following writings are equivalent:
[tex] (g\circ h \circ f)(x) = g(h(f(x))) [/tex]
Using the second form, we can see that the first thing to do is to compute [tex] f(x) [/tex], which is given:
[tex] f(x) = x-3 [/tex]
This means that we can update the fomula:
[tex] (g\circ h \circ f)(x) = g(h(f(x))) = g(h(x-3)) [/tex]
So, now we have to compute [tex] h(x-3) [/tex]. Given [tex] h(x)=2x+1 [/tex], we know that [tex] h[/tex] doubles the input and adds one. In this case, the input is [tex] x-3 [/tex], so we have to double this and add one:
[tex] h(x-3) = 2(x-3)+1 = 2x-6+1 = 2x-5 [/tex]
So, we have
[tex] (g\circ h \circ f)(x) = g(h(f(x))) = g(2x-5) [/tex]
Similarly, from [tex] g(x) = x+3 [/tex], we understand that [tex] g [/tex] just adds 3 to the input. In this case, the input is [tex] 2x-5 [/tex], so we'll add 3 to this quantity:
[tex] g(2x-5) = (2x-5)+3 = 2x-2 [/tex]
So, we finally have
[tex] (g\circ h \circ f)(x) = g(h(f(x))) = 2x-2 [/tex]
