Answer:
[tex]x=1[/tex] is a critical point of the function [tex]f(x).[/tex]
Step-by-step explanation:
[tex]f(x)=x^2-2x+7[/tex]
Differentiate both sides w.r.t. [tex]x[/tex]
[tex]\Rightarrow f'(x)=2x-2[/tex]
Now, [tex]f'(x)=0[/tex]
[tex]\therefore 2x-2=0[/tex]
[tex]\Rightarrow 2x=2[/tex]
[tex]\Rightarrow x=1[/tex]
Hence, [tex]x=1[/tex] is a critical point of the given function.
