The First Derivative Test

# 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)
\begin{align} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x = \frac{4 \pm \sqrt{4}}{6} \\ x = \frac{1}{3} \: , \: 1 \end{align}

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.