First derivative test

First derivative test. A function $f(x)$ has a maxima at $x=c$ if:

1. $f'(c)=0$,
2. $f'(c-h)>0$ [$f(x)$ is increasing for values of $x<c$], and
3. $f'(c+h)<0$ [$f(x)$ is decreasing for values of $x>c$], where $h$ is a small positive number. A function $f(x)$ has a minima at $x=c$ if:

(i) $f'(c)=0$,

(ii) $f'(c-h)<0$ [$f(x)$ is decreasing for values of $x<c$], and

(iii) $f'(c+h)>0$ [$f(x)$ is increasing for values of $x>c$].

Note. If $f'(c)=0$ but $f'(c-h)$ and $f'(c+h)$ have the SAME sign (both positive, or both negative), then $x=c$ is neither a maxima nor a minima — such a point is called a point of inflexion. Examples include $f(x)=x^3$ and $f(x)=x^5$ on $[-2,2]$ at $x=0$, where the derivative is momentarily zero but the function keeps increasing right through that point.

Worked Example. Find the local maxima or local minima of $f(x)=x^3-3x$. Differentiating: $f'(x)=3x^2-3=3(x^2-1)$. Setting $f'(x)=0$: $x^2=1\Rightarrow x=\pm1$ — these are the turning points.

At $x=1$: taking $x=1-h$ for small $h>0$, $f'(1-h)=3[(1-h)^2-1]=3h(h-2)$, which is negative (since $h>0$ and $h-2<0$) — so $f'$ is negative just to the left of $x=1$ (the function is decreasing there). Taking $x=1+h$: $f'(1+h)=3[(1+h)^2-1]=3(h^2+2h)$, which is positive — so $f'$ is positive just to the right of $x=1$. Since $f'$ goes from negative to positive through $x=1$, this is a point of local minima: $f(1)=1-3=-2$. …