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x=0 or y=0Surely they say something about the domain of the original function? They don't consider 0 a critical point of 1/x do they?
This. Take first derivative. then solve x for dy/dx = 0The syllabus definition is when the first derivative is undefined or zero.
Previous post edited for your sake.Yes, after a comment like a definition is useless, only to follow by a (not so helpful) definition, you're missing an invitation to leave the mathematics forum for the mathematical crimes of (1) disparaging definitions, and (2) abuse of logic
I'm happy to oblige.
Joking .. mostly
Well no, because the second derivative would not be undefined or zero, as per the definition Trebla provided.Cmon. Is -2 a critical point of sqrt(x)?
Implicit here is that x is in the domain of f.Um i think the definition of a function should be raised.
From Wolfram:
A function is a relation that uniquely associates members of one set with members of another set.
A function y=f(x) has critical points at all points x0 where f^'(x0)=0 or f(x) is not differentiable.
Nice.So f: R -> R (real numbers in domain mapping to real numbers in codomain), where f(x)=1/x is not a function as f(0) is not defined. Thus f does not uniquely associate 0 with a real number in the codomain and f is not a function, thus we cannot use the definition of critical points for a function on f. Thus we cannot say 0 is a critical point.
However, if f: R/{0} -> R (real numbers excluding 0 in domain mapping to real numbers in codomain) where f(x)=1/x is a function, as every real number in the domain can be mapped to a real number in the codomain. Thus the definition of critical points for a function can be applied. However, 0 is not in the domain of f, so it cannot possibly be a critical point.
Your function h is not well-defined without choosing a branch of sqrt map. Otherwise, nice (assuming you apply the same observation about the domain/critical points as you did above).Also, for f: R--> R, f(x)=sqrt(x) is not a function, as it does not map negative numbers to any numbers in the real codomain, however for g: (0, inf) --> R, g(x)=sqrt(x) and h:Real --> Complex where h(x)=sqrt(x) are functions.
They definitely aren't useless, and definitions can't be wrong - they're definitions. They should be consistent with what is likely to be seen in later work though.EDIT: Yeh but for high school definitions are useless, and sometimes wrong...