Question 162319
Yes, you substitute g(x) for x in f(x).
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{{{f(x)=x^2-2x+1}}}
{{{g(x)=sqrt(x-2)}}}
{{{g^2(x)=(x-2)}}}
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{{{f(g(x))=g^2-2g+1}}}
{{{f(g(x))=(x-2)-2*sqrt(x-2)+1}}}
{{{f(g(x))=x-2*sqrt(x-2)-1}}}
As you see you just multiply by the square root and leave it as is.