Question 6072

f(x+1) for {{{sqrt (x^2+x)}}}

={{{sqrt( (x+1)^2+X+1)}}}
={{{sqrt( x^2+2x+1+x+1)}}}
={{{sqrt (x^2+3x+2)}}}

You were correct up to here:
={{{2*sqrt(3x+1)}}}
the 2 does not come out of the square root, so your answer should have stayed ={{{sqrt( x^2+3x+2)}}}