Q:

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

Accepted Solution

A:
Answer:[tex]\large\boxed{B.\ 4x^2-41x+105}[/tex]Step-by-step explanation:[tex]f(x)=4x-\sqrt{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-\sqrt{(x-5)^2}\\\\\text{use}\\(a-b)^2=a^2-2ab+b^2\\\sqrt{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[/tex]