How do you find (f*g)(-1)(fg)(1) given f(x)=x^2-1f(x)=x21 and g(x)=2x-3g(x)=2x3 and h(x)=1-4xh(x)=14x?

1 Answer
Sep 12, 2017

(f * g)(-1)=24(fg)(1)=24

Explanation:

I personally find it easier to think of
color(white)("XXX")(f * g)(x)XXX(fg)(x) and (f * g)(-1)(fg)(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)=x21
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))21

Now since we are also given that
color(white)("XXX")color(blue)(g(color(magenta)x)=2color(magenta)x-3XXXg(x)=2x3
after substituting color(blue)(2color(magenta)x-3)2x3 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))=(2x3)21

Expanding (and dropping the color coding for a bit)
color(white)("XXX")f(g(x))=4x^2-12x+8XXXf(g(x))=4x212x+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))=4x212x+8

Since we are asked for (f * g)(color(red)(-1))(fg)(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)212(1)+8

color(white)("XXXXXXXX")=4+12+8XXXXXXXX=4+12+8

color(white)("XXXXXXXX")=24XXXXXXXX=24