Question

In the picture I need the answers to the questions.

Answer 1

Answer:

a) $\mathbf{ f(g(x))=x+2\sqrt{x} +1}$

b) $\mathbf{f(g(9))=16}$

Step-by-step explanation:

We are given:

$f(x)=(x-2)^2\\g(x)=\sqrt{x} +3$

We need to find  $a) f(g(x))\\b) f(g(9))$

a) First finding: $f(g(x))$

It can be found by putting the value of g(x) into f(x)

We are given:

$f(x)=(x-2)^2\\Put\:x=g(x)\:i.e. \: x= \sqrt{x} +3\\f(g(x))=(\sqrt{x} +3-2)^2\\Now\: solving:\\f(g(x))=(\sqrt{x} +1)^2\\Using\:formula\:\mathbf{(a+b)^2=a^2+2ab+b^2}\\f(g(x))=(\sqrt{x})^2+2(\sqrt{x} )(1)+(1)^2\\ f(g(x))=x+2\sqrt{x} +1$

SO, we get: $\mathbf{ f(g(x))=x+2\sqrt{x} +1}$

b) Now finding: $f(g(9))$

It can be found by putting x=9 in f(g(x))

We have:

$f(g(x))=x+2\sqrt{x} +1\\Put\:x=9\\f(g(9))=9+2\sqrt{9}+1\\We\:know\: \sqrt{9}=3\\ f(g(9))=9+2(3)+1\\f(g(9))=9+6+1\\f(g(9))=16$

So, we get: $\mathbf{f(g(9))=16}$