I personally find it easier to think of
color(white)("XXX")(f * g)(x)XXX(f⋅g)(x) and (f * g)(-1)(f⋅g)(−1)
as f(g(x))f(g(x)) and f(g(-1)f(g(−1) respectively
Given
color(white)("XXX")f(color(blue)x)=color(blue)x^2-1XXXf(x)=x2−1
then
after simply substituting color(blue)(g(x))g(x) for xx
color(white)("XXX")f(color(blue)(g(x)))=(color(blue)(g(x)))^2-1XXXf(g(x))=(g(x))2−1
Now since we are also given that
color(white)("XXX")color(blue)(g(color(magenta)x)=2color(magenta)x-3XXXg(x)=2x−3
after substituting color(blue)(2color(magenta)x-3)2x−3 for color(blue)(g(x))g(x) above
we get
color(white)("XXX")f(color(blue)(g(color(magenta)x)))=(color(blue)(2color(magenta)x-3))^2-1XXXf(g(x))=(2x−3)2−1
Expanding (and dropping the color coding for a bit)
color(white)("XXX")f(g(x))=4x^2-12x+8XXXf(g(x))=4x2−12x+8
...or we could re-color code the variable color(red)xx to make this appear as
color(white)("XXX")f(g(color(red)(x)))=4color(red)x^2-12color(red)x+8XXXf(g(x))=4x2−12x+8
Since we are asked for (f * g)(color(red)(-1))(f⋅g)(−1)
which can be written as f(g(color(red)(-1)))f(g(−1))
we can substitute (color(red)(-1))(−1) in place of color(red)xx in our definition of f(g(color(red)x))f(g(x))
color(white)("XXX")f(g(color(red)(-1)))=4 * (color(red)(-1))^2-12 * (color(red)(-1))+8XXXf(g(−1))=4⋅(−1)2−12⋅(−1)+8
color(white)("XXXXXXXX")=4+12+8XXXXXXXX=4+12+8
color(white)("XXXXXXXX")=24XXXXXXXX=24