Question

Given that f(x) = x2 − 4x − 3 and g(x) = the quantity of x plus three, over four , solve for f(g(x)) when x = 9? −6 3 6 9

Given that `f(x) = x2 − 4x − 3` and `g(x) =( x+3)/4`, solve for `f(g(x))` when `x = 9`?

`6 ; 3; 6; 9`

Answer 1

`-6`

Explanation:

`"to evaluate "f(g(x))" substitute "x=g(x)" into "f(x)`

`g(x)=(x+3)/4`

`f(g(x))=f((x+3)/4)`

`=((x+3)/4)^2-cancel(4)((x+3)/cancel(4))-3`

`=((x+3)^2)/16-(x+3)-3`

`f(g(color(red)(9)))=((color(red)(9)+3)^2)/16-(color(red)(9)+3)-3`

`color(white)(xxxxx)=144/16-12-3`

`color(white)(xxxxx)=9-15=-6`

Answer 2

I will only do the first one for you. It has a full explanation

`f(g(9))=-6`

Explanation:

Given that `g(color(magenta)(x))=(color(magenta)(x)+3)/4 color(white)("ddd")->color(white)("ddd") g(color(magenta)(9)) = (color(magenta)(9)+3)/4`

So wherever there is an `x` put a `color(magenta)(9)` instead.

`g(9)=12/4=color(green)(3)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that `f(x)= x^2-4x-3`

So wherever there is a `g(x)` put a `color(green)(3)`

`f(g(x)) =f(color(white)("d")ubrace(g(color(magenta)(9)))color(white)("d"))`
`color(white)("dddddddddddd")darr`
`f(g(x)) = f(color(white)("d.d")color(green)(3)color(white)("dd."))`

`color(brown)(f(color(green)(3))=color(green)(3)^2color(white)("dd")ubrace(-4(color(green)(3))-3))`
`color(brown)(color(white)("ddddddddddd.d")darr`
`f(3)=+9color(white)("d")-15color(white)("ddd")=-6`