First Derivative Test
Calculus & Advanced MathThe first derivative test classifies a critical point as a local maximum, local minimum, or neither by checking whether the derivative changes sign there.
Definition
After finding a critical point, the first derivative test checks the sign of the derivative just before and just after that point. If the sign goes from + to -, it is a local max. If it goes from - to +, it is a local min.
Example
$f(x) = x^2 - 4$: $f'(x) = 2x = 0$ at $x = 0$. For $x < 0$, $f' < 0$ (falling). For $x > 0$, $f' > 0$ (rising). Sign goes $-$ to $+$, so $x = 0$ is a local minimum.
Key Insight
Think of hiking: if you were going downhill and then start going uphill, you just passed through a valley (local min). Downhill then uphill means you hit a peak (local max).
Definition
At a critical point $c$: if $f'$ changes from positive to negative, $c$ is a local max. If $f'$ changes from negative to positive, $c$ is a local min. If $f'$ does not change sign, $c$ is neither (possibly an inflection point).
Example
$f(x) = x^3 - 3x$: critical points at $x = \pm 1$. At $x = -1$: $f'$ goes $+$ to $-$, so local max $f(-1) = 2$. At $x = 1$: $f'$ goes $-$ to $+$, so local min $f(1) = -2$.
Key Insight
The first derivative test is more versatile than the second derivative test because it works even when $f'' = 0$ at the critical point.
Definition
The first derivative test is a sign-analysis technique requiring only the sign of $f'$ near $c$, not its magnitude. It is applicable even when $f''$ does not exist. It follows directly from the relationship between monotonicity and the sign of $f'$ established by the Mean Value Theorem.
Example
$f(x) = x^{1/3}$: not differentiable at $0$, but $f$ changes from decreasing ($f' < 0$ for $x < 0$) to increasing ($f' > 0$ for $x > 0$)... wait, $f(x) = x^{1/3}$ is actually increasing everywhere. Instead: $f(x) = x^{2/3}$ has a local min at $0$ by sign analysis even though $f'(0)$ is undefined.
Key Insight
In variational calculus, the analogue of the first derivative test is the Euler-Lagrange equation: a functional has a critical point where its "derivative" (functional derivative) vanishes.