1. Apr 7, 2014

### LagrangeEuler

Is it necessary for some point $x$ of the function to be saddle that
$f'(x)=0$?

Last edited: Apr 7, 2014
2. Apr 7, 2014

### Staff: Mentor

What you have written isn't clear. For one thing, saddle points aren't applicable to functions of a single variable. For another, do you mean that f'(x) = 0 for some specific value of x? Or do you mean that f'(x) is identically equal to zero?

If you meant f'(p) = 0, for some p, all this means is that at the point (p, f(p)), the tangent to the curve is horizontal. The following three functions all have horizontal tangents when x = 0.
1. f(x) = x2 - there is a local (and global) minimum for x = 0.
2. g(x) = x3 - there is an inflection point for x = 0. This function has no minimum and no maximum.
3. h(x) = 1 - x2 - there is a local (and global) maximum for x = 0.

If you meant f'(x) $\equiv$ 0 (i.e., identically equal to zero), it must be the case that f(x) = C, the graph of which is a horizontal line. This "curve" has no minimum and no maximum.

3. Apr 7, 2014

### LagrangeEuler

Yes I mean for particular value of $x$. For some $x_0$ is it possible situation that
$f'(x_0)\neq 0$ and that in $x_0$ function has inflection point?

4. Apr 7, 2014

### Staff: Mentor

Yes. Let f(x) = x1/3.
f' is not defined at x = 0, nor is f'', but the curvature changes from concave up for x < 0 to concave down for x > 0. The fact that the curvature changes direction on either side of x = 0 is sufficient to be able to state that there is an inflection point for x = 0, which is in the domain of this function.