Answer:
No
Step-by-step explanation:
Given
f(x) = [tex]\sqrt{x-10}[/tex] + 11 ← substitute x = 1
f(1) = [tex]\sqrt{1-10}[/tex] + 11
= [tex]\sqrt{-9}[/tex] + 11
[tex]\sqrt{-9}[/tex] is not real and therefore x = 1 does not have a real output for f(x)
