# Fixed points and graphs

1. Mar 27, 2006

### Benny

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.

2. Mar 27, 2006

### nocturnal

if f(x) = x, then f'(x) = 1 for all x. so x=0 is not a critical point. f is an increasing function.

Last edited: Mar 27, 2006
3. Mar 27, 2006

### HallsofIvy

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.

4. Mar 29, 2006

### Benny

Oops, I should watch my differentiation...better hope that doesn't happen during an exam.