Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

Answer 2

Answer: Option B

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

Step-by-step explanation:

We have 2 functions

$f(x) = 4x -\sqrt{x}$

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

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

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

$f(g(x)) = 4(g(x)) -\sqrt{(g(x))}$

$f(g(x)) = 4(x-5)^2 -\sqrt{(x-5)^2}$

By definition $\sqrt{a^2} = |a|$

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

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

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

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

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