Fixed Points and Critical Points in Continuous Scalar Functions

AI Thread Summary
The discussion centers on the nature of critical points in continuous scalar functions, specifically addressing the example of f(x) = x in one dimension. It is clarified that x = 0 is not a critical point because the derivative f'(x) = 1 is non-zero for all x. The conversation also touches on the concept of fixed points, questioning whether a point on a line of fixed points can correspond to a maximum or minimum. A participant corrects the initial assumption, suggesting that a different function, such as f(x) = (1/2)x^2 + C, would indeed have a minimum at x = 0. Overall, the thread emphasizes the importance of correctly identifying critical points and their implications in higher dimensions.
Benny
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Hi, I was thinking about the following and would like some clarification. Suppose that we have a continuous scalar function f:R^n \to R with a critical point at say x_0, where the dimension of x_0 depends on the value of n.

Consider as an example f(x) = x (n = 1). The point x = 0 is a critical point since f'(x) is zero at that point. Since f is continuous then corresponding to x = 0 must be a local minimum, local maximum or saddle correct? (Not exactly sure about it)

My point is that x = 0 lies on a line of fixed points and hence cannot correspond to a maximum or a minimum? Is this true in higher dimensions or does this reasoning hold at all?

Any help would be good thanks.
 
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Benny said:
Hi, I was thinking about the following and would like some clarification. Suppose that we have a continuous scalar function f:R^n \to R with a critical point at say x_0, where the dimension of x_0 depends on the value of n.

Consider as an example f(x) = x (n = 1). The point x = 0 is a critical point since f'(x) is zero at that point.
if f(x) = x, then f'(x) = 1 for all x. so x=0 is not a critical point. f is an increasing function.
 
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As nocturnal said, if f(x)= x then x= 0 is NOT a critical point. Maybe you were thinking of f'(x)= x which would correspond to f(x)= (1/2)x2+ C. That really does have a minimum at x= 0.
 
Oops, I should watch my differentiation...better hope that doesn't happen during an exam.:biggrin:
 
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