First, let's see what #(f @ g)(x)# means.
#(f @ g)(x)#, or perhaps simply #f @ g(x)#, is shorthand for #f(g(x))#. That means shoving in a value for #x#, solving for #g(x)# by shoving in that value of #x#, then shoving in that solved value into the function #f#, hence #f(g(x))#.
#f @ g(3)# is then simply #f(g(3))#. We're already given the value of #x#, which is #3#, and we're told to solve this function composition: first solving for #g(3)#, then solving for the function #f# having shoved in the value of #g(3)#.
Here, we have #f(x) = -2x + 8# and #g(x) = 5x#. Since an equation, any equation, tells us that both sides are equal (and can be replaced by each other), why don't we replace #g(x)# with #5x#?
Doing that, #f(g(x))# becomes #f(5x)#, so #f(g(x)) = f(5x)#.
Since #f(x) = -2x + 8# for any input #x# (which means we might as well say #f(n) = -2n + 8# and it would still be the same), solving for #f(5x)# would simply be shoving in that #5x# as the input:
#f(5x) = -2(5x) + 8#
Simplify:
#f(5x) = -10x + 8#
Summarizing what we've done so far, we have:
#(f @ g)(x) = f @ g(x) = f(g(x)) = f(5x) = -10x + 8#
Now, we just need to solve for #x = 3#:
#(f @ g)(3) = -10(3) + 8#
# = -30 + 8 = -22#
Therefore, #(f @ g)(3) = -22#.
