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If $f'(x)=0$ and $f''(x)\neq0$, does it mean that function $f$ has a local extremum in $x$?
If $f'(x)=0$ and $f''(x)=0$, does it mean that function $f$ has no local extremum in $x$?

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    $\begingroup$ Why don't you start with some examples like $f(x) = x^2 ; x^3$ which may gives you some insight into a more formal solution. $\endgroup$ – Ram Feb 10 '13 at 12:39
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If $f'(x)=0$ and $f''(x)\neq0$, then there is a local extremum in $x$. When $f''(x)=0$, there is a horizontal slope and an inflection point, and not necessarily an extremum. Think of $f(x)=x^3$. As @DavidMitra points out in the comment below, this reasoning applies only to odd powers. For even powers, there is a local extremum.

For surfaces in two dimensions, places where the gradient vanishes, as well in places where either the second partial derivatives vanish or a matrix of second derivatives (known as a Hessian) is indefinite is called a saddle point.

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  • $\begingroup$ What about $f(x)=x^4$? This does not have an inflection point at $x=0$, it does have an extremum at $x=0$, yet $f'(x)=f''(x)=0$. $\endgroup$ – David Mitra Feb 10 '13 at 13:35
  • $\begingroup$ @DavidMitra: In my haste, I did not add that my reasoning applies only to odd powers. Thanks. $\endgroup$ – Ron Gordon Feb 10 '13 at 13:37

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