The Second Derivative Test

# The Second Derivative Test

We have already looked at one way to determine whether a critical number $c$ is a local maximum, minimum, or neither by the The First Derivative Test. Sometimes it may be difficult to compute values to the left and right of a critical number though, so we will look at another method involving the second derivative of a function $f$.

 Strategy (The Second Derivative Test): Let $f''$ is a continuous function at let $c$ be a critical number such that $f'(c) = 0$. If $f''(c) > 0$, then the point $(c, f(c))$ is a local minimum. If $f''(c) < 0$, then the point $(c, f(c))$ is a local maximum.

To understand the second derivative test, it is important to recall that a derivative tells us the rate of change. Therefore, if $f''(c) > 0$, then at $(c, f(c))$, we know that the derivative of $f$ is increasing - that is the slopes are increasing. If the slopes around critical number $c$, then $(c, f(c))$ must be a local minimum. Similarly, if $f'(c) < 0$, then at $(c, f(c))$, we know that the derivative of $f$ is decreasing - that is the slopes are decreasing. If the slopes around critical number $c$ are decreasing, then $(c, f(c))$ must be a local maximum.

We will now use the second derivative test in a couple of examples.

## Example 1

Determine and classify the critical points of the function $f(x) = \sin x$ over the interval $[0, 2\pi]$.

We must first differentiate $f$ to get $f'(x) = \cos x$. To find critical numbers $c$, let $f'(x) = 0$. Note that $\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

Now let's differentiate $f'$ to get $f''(x) = -\sin x$. Inputting $x = \frac{\pi}{2}$, we get $f''(\frac{\pi}{2}) = -1$, and therefore, $( \frac{\pi}{2}, 1)$ is a local maximum.

Furthermore, inputting $x = \frac{3\pi}{2}$, we get $f''(\frac{3\pi}{2}) = 1$, and therefore, $( \frac{3\pi}{2}, -1)$ is a local minimum. This can easily be verified by looking at the graph of $f(x) = \sin x$:

## Example 2

Determine and classify the critical points of the function $f(x) = -x^4 + x^2$.

If we differentiate $f$ we get $f'(x) = -4x^3 + 2x$. We thus have critical points at $f'(x) = 0$:

(1)
\begin{align} 0 = -4x^3 + 2x \\ 0 = -2x(2x^2 -1) \end{align}

So $f'(x) = 0$ when $x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$. If we differentiate $f'$, we get that $f''(x) = -12x^2 + 2$, and inputting our critical numbers in:

$f''(0) = 2$ $f''(\frac{1}{\sqrt{2}}) = \frac{1}{4}$ $f''(-\frac{1}{\sqrt{2}}) = \frac{1}{4}$

So therefore by the second derivative test, $(0, 0)$ is a local minimum, and $(-\frac{1}{\sqrt{2}}, \frac{1}{4})$ and $(\frac{1}{\sqrt{2}}, \frac{1}{4})$ are local maximums.