The First Derivative Test
The first derivative test is a strategy that allows us to determine whether a critical number $c$ is either an extrema (maximum or minimum), or neither. It is important to be familiar with Concavity and Inflection Points as well as extreme values before reading forward.
| Strategy (The First Derivative Test): Let $c$ be a critical number of the continuous function $f$. a) If $f' > 0$ at the left of $c$ and $f' < 0$ at the right of $c$, then $(c, f(c))$ is a local maxima. b) If $f' < 0$ at the left of $c$, and $f' > 0$ at the right of $c$, then $(c, f(c))$ is a local minima. _ c) If $f'$ does not change signs on the left and right of $c$, then $(c, f(c))$ is neither a local maxima or local minima. |
In other words, if the sign of the derivative of $f$ changes on either side of the point $(c, f(c))$, then this point is either a local maxima or minima, and if the sign doesn't change, then it is neither. We will now look at an example.
Example 1
Show that $f(x) = 2x^2$ has a local minimum at $(0, 0)$.
When we differentiate $f$, we get $f'(x) = 4x$, and setting $f'(x) = 0$, we have one critical number $c = 0$.
Now let's test values to the left of $0$, for example, $-1$. Clearly, $f'(-1) = -4 < 0$. If we test values to the right of $0$, for example $1$, we get that $f(1) = 4 > 0$. Therefore, $f' < 0$ to the left of $0$, and $f' > 0$ to the right of $0$, so $(0, 0)$ must be a local minimum by the first derivative test.
Example 2
Find and classify all critical numbers for the function $f(x) = x^3 - 2x^2 + x$.
We first differentiate $f$ to get $f'(x) = 3x^2 -4x + 1$, and then set $f'(x) = 0$. Using the quadratic formula, we obtain:
(1)Let's now set up a chart and test values to the left and right of both critical points:
| $f'(0)$ | $f'(\frac{1}{3})$ | $f'(\frac{1}{2})$ | $f'(1)$ | f'(2) | |
|---|---|---|---|---|---|
| Sign of f' | $+$ | $0$ | $-$ | $0$ | $+$ |
By the first derivative test, $(\frac{1}{3}, f(\frac{1}{3}))$ is a local maximum, and $(1, f(1))$ is a local minimum.