How do you find #(f*g)(-1)# given #f(x)=x^2-1# and #g(x)=2x-3# and #h(x)=1-4x#?

Question

Hazel Anglin · Accepted Answer

undefined (f * g)(-1) = (f(-1)) * (g(-1))
f(-1) = (-1)^2 - 1 = 0
g(-1) = 2 * (-1) - 3 = -5
(f * g)(-1) = 0 * (-5) = 0

Eli Assouline · Answer

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