Question

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

Answer 1

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

Explanation:

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`