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The Second Derivative Test (for Local Extrema)

In addition to the first derivative test, the second derivative can also be used to determine if and where a function has a local minimum or local maximum.

Consider the situation where $c$ is some critical value of $f$ in some open interval $(a,b)$ with $f'(c)=0$.

Let us also presume that $f$ is differentiable in this open interval and that $f''(c)$ exists.

Clearly, the second derivative must be either negative, zero, or positive.

Let us explore the first case, where $f''(c) = L$, with $L \lt 0$. By the limit definition of the derivative, we have

By the epsilon-delta definition of a limit, we know that for any $\epsilon \gt 0$ we can find a $\delta \gt 0$ so that for any $x$ in the open interval $(c-\delta,c+\delta)$ except maybe $x=c$, the value of $f(x)$ is in the interval $(L-\epsilon,L+\epsilon)$.

Choosing $\epsilon = |\frac{L}{2}|$, let us more simply denote this $\delta$-interval about $c$ as $(a,b)$.

Keeping in mind that if one travels no more than $|\frac{L}{2}|$ from $L$, which is negative -- one must still be negative -- we have thus found an open interval $(a,b)$ containing $c$ with

Then for any $x$ in $(a,c)$, we have $x-c \lt 0$. Combined with the inequality above, this tells us $f'(x) - f'(c) \gt 0$. Recalling $f'(c)=0$, we have $f'(x) \gt 0$ for every $x$ in $(a,c)$.

Similarly, for any $x$ in $(c,b)$, we have $x-c \gt 0$. Again combining this with the previous inequality tells us $f'(x) - f'(c) \lt 0$. Again recalling $f'(c) = 0$, we have $f'(x) \lt 0$ for every $x$ in $(c,b)$.

Of course, we can now apply the first-derivative test and conclude that $f$ has a local maximum at $c$.

The case where $f''(c) \gt 0$ can be argued similarly to reveal a local minimum at $c$. The following result summarizes the conclusions just made: