Question

What is f(g(x)) for x > 5?

Answer 1

`\large\boxed{B.\ 4x^2-41x+105}`

Explanation:

`f(x)=4x-√(x)\\\\g(x)=(x-5)^2\\\\f(g(x))\to\text{put}\ x=(x-5)^2\ \text{to}\ f(x):\\\\f(g(x))=f\bigg((x-5)^2\bigg)=4(x-5)^2-√((x-5)^2)\\\\\text{use}\\(a-b)^2=a^2-2ab+b^2\\√(x^2)=|x|\\\\f(g(x)=4(x^2-2(x)(5)+5^2)-|x-5|\\\\x>5,\ \text{therefore}\ x-5>0\to|x-5|=x-5\\\\f(g(x))=4(x^2-10x+25)-(x-5)\\\\\text{use the distributive property:}\ a(b+c)=ab+ac\\\\f(g(x))=(4)(x^2)+(4)(-10x)+(4)(25)-x-(-5)\\\\f(g(x))=4x^2-40x+100-x+5\\\\\text{combine like terms}\\\\f(g(x))=4x^2+(-40x-x)+(100+5)\\\\f(g(x))=4x^2-41x+105`

Answer 2

Answer: Option B

`f(g(x)) = 4x^2 -41x + 105`

Explanation:

We have 2 functions

`f(x) = 4x -√(x)`

`g(x) = (x-5)^2`

We must find
`f(g(x))`

To find this composite function enter the function g(x) within the function f(x) as follows

`f(g(x)) = 4(g(x)) -√((g(x)))`

`f(g(x)) = 4(x-5)^2 -√((x-5)^2)`

By definition
`√(a^2) = |a|`

So

`f(g(x)) = 4(x-5)^2 -|x-5|`

Since x is greater than 5 then the expression
`(x-5)> 0`.

Therefore we can eliminate the absolute value bars

`f(g(x)) = 4(x-5)^2 -(x-5)`

`f(g(x)) = 4(x^2 -10x + 25) -(x-5)`

`f(g(x)) = 4x^2 -40x + 100 -x+5`

`f(g(x)) = 4x^2 -41x + 105`