Using the Second Derivative Test to Determine Extrema

#5.7 Using the Second Derivative Test to Determine Extrema

You’ve probably noticed by now that Unit 5 deals with analytical applications of differentiation; that means that a function’s derivatives can tell us something about its behaviors. We learned from 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema that the first derivative of a function tells us information on where its relative extrema (aka local maximum and minimum) will be.

What does the second derivative tell us, then? Answer: Whether a critical point we found yields a local maximum or minimum (it can’t be both)! 🧠

#✏️ Warm-up: Finding Critical Points

Let’s do a quick recap of critical points from key topic 5.2! To refresh, a critical point is a point on a function where…

• the first derivative equals zero or
• fails to exist Analytically, the first bullet point is easier to wrap your head around. Let’s take a look at a quick example:

$$f(x)=\frac{2}{3}x^3-\frac{5}{2}x^2-3x$$

Computing the first derivative gives us:

$$f'(x)=2x^{2}-5x-3$$

To find where the first derivative equals zero, we can factor and then set f’(x) = 0:

$$0=2x^2-5x-3$$

$$0=(2x+1)(x-3)$$

Looks familiar, right? We proceed as if we’re finding the roots (x) in a regular equation. In this case, the two x values are our critical points!

$$x=-\frac{1}{2}, x=3$$

Not bad, right? What do we do with these critical point values, then?

We learned two things from 5.6 Determining Concavity of Functions over Their Domains about what the second derivative of a function can tell us information of:

1. 🌈 The function’s intervals of upward or downward concavity.
2. $f’’(...