Properties of second derivative to first derivative
I know that if the second derivative of f is positive, and the first derivative of f exists, then the point of the first derivative is a local minimum, and if the second derivative of f is negative then, the point of the first derivative f is a maximum. But if the second derivative of f is exactly 0, then what does that tell us?
Thanks
$\endgroup$ 13 Answers
$\begingroup$What you mean to say is that suppose $f'(c) = 0$.
If $f''(c) > 0$, then $f$ has a local minimum at $c$.
If $f''(c) < 0$, then $f$ has a local maximum at $c$.
However, if $f''(c) = 0$, then we get no information.
This is commonly referred to as the Second Derivative Test.
Let's take a look at the $f''(c) = 0$ case.
For instance, if $f(x) = x^4$. Then $f'(0) = f''(0) = 0$ and we happen to have a local minimum.
But if $f(x) = -x^4$, then again $f'(0) = f''(0) = 0$, but this time we have a local maximum.
You could also have neither. Let $f(x) = x^3$, $f'(0) = f''(0) = 0$, but $f$ has neither a local max nor min at $0$.
$\endgroup$ $\begingroup$The more general statement is this:
Consider the first nonzero derivative $\frac{d^n f}{dx^n}$, where $n>0$. If $n$ is odd, then the function has neither a minimum nor a maximum at the point. If $n$ is even and $\frac{d^nf}{dx^n}>0$, then the point is a local minimum. If $n$ is even and $\frac{d^nf}{dx^n}<0$, then the point is a local maximum.
This assumes that $f$ is smooth - i.e. that its derivatives exist for all $n$. If that isn't the case, you need to be more careful, but this assumption is part of the 2nd derivative test anyway.
$\endgroup$ 1 $\begingroup$In general, when derivatives exist:
if $f'(x_0)=0$ and $\exists k \geq 2$ s.t. $f^k(x_0) \neq 0$ then
when k is even you have a max/min in $x_0$ (depending on the sign)
when k is odd you have an inflection point in $x_0$
