Cracking the Second Derivative Test: AP Calculus AB/BC Study Guide
Introduction
Hello, all you calculus adventurers! Get ready to dive into the wild world of the Second Derivative Test. If you thought the First Derivative Test was magical, wait till you see what its younger sibling, the Second Derivative Test, can do! 🌈🧙♂️
What the Second Derivative Really Tells You
Remember the First Derivative Test from Unit 5.4? It gave us clues about where functions hit their local high points (maximums) and low points (minimums). The Second Derivative Test, like Sherlock Holmes for calculus, helps us determine if a critical point (a point where the first derivative is zero or undefined) is a local maximum or a local minimum—and it does this in style! 🕵️♂️
But how, you ask? Simple! The second derivative (basically the derivative of the first derivative) gives us the concavity of the function: whether the graph is curvy like a bowl 🍜 (concave up) or hilllike ⛰️ (concave down).
Before we go deeper, let’s shake off the cobwebs with a quick recap on finding critical points.
WarmUp: Finding Critical Points
Critical points are like plot twists in your favorite TV show. They occur where the first derivative of a function is either zero or doesn’t exist (because math likes to keep things interesting).
Take, for example, the function: [ f(x) = \frac{2}{3}x^3  \frac{5}{2}x^2  3x ]
First, compute the first derivative: [ f'(x) = 2x^2  5x  3 ]
To find those elusive critical points, solve for where ( f'(x) = 0 ): [ 0 = 2x^2  5x  3 ] [ 0 = (2x + 1)(x  3) ]
Voilà! The critical points are: [ x = \frac{1}{2}, \quad x = 3 ]
Playing with Concavity: What the Second Derivative Reveals
The second derivative is our trusty changeometer, telling us where the function’s graph smiles (concave up) or frowns (concave down).
For our function ( f(x) ): [ f''(x) = 4x  5 ]
Steps for the Second Derivative Test

Find the Critical Points (Already Did That!) We have ( x = \frac{1}{2} ) and ( x = 3 ).

Evaluate the Second Derivative at each Critical Point Plug the critical points into ( f''(x) = 4x  5 ):
[ f''(\frac{1}{2}) = 4(\frac{1}{2})  5 = 7 \quad \text{(yikes, it’s negative!)} ] [ f''(3) = 4 \cdot 3  5 = 7 \quad \text{(hooray, it’s positive!)} ]

Determine Local Minima and Maxima If ( f''(x) > 0 ), the function is concave up, and the critical point is a local minimum (think basin of a bowl 🍜). If ( f''(x) < 0 ), the function is concave down, and the critical point is a local maximum (think peak of a hill ⛰️).
So:
 ((0.5, 7)) is a local maximum.
 ((3, 7)) is a local minimum.
Why It Makes Sense
Visualize the maximum as standing proudly at the top of a hill, surveying the lands—or that stubborn hair on your head that just won’t lie flat! 🗻 Conversely, picture the minimum as the deepest part of a cauldron, where all the tasty soup gathers. 🥣
Practice Makes Perfect!
Now it’s your turn to get your hands dirty. Here are a couple of functions to test your newfound skills on:
Example 1
[ f(x) = 4 \sin(x), \quad 0 < x < 2\pi ]
First find the derivatives: [ f'(x) = 4 \cos(x) ] [ f''(x) = 4 \sin(x) ]
Set ( f'(x) = 0 ): [ \cos(x) = 0 ] [ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} ]
Now plug these into ( f''(x) ): [ f''\left(\frac{\pi}{2}\right) = 4 \sin\left(\frac{\pi}{2}\right) = 4 \quad (\text{local max}) ] [ f''\left(\frac{3\pi}{2}\right) = 4 \sin\left(\frac{3\pi}{2}\right) = 4 \quad (\text{local min}) ]
Example 2
[ f(x) = 247\ln(x^2) ]
Calculate the derivatives: [ f'(x) = \frac{247}{x} ] [ f''(x) = \frac{247}{x^2} ]
Set ( f'(x) = 0 ): [ \frac{247}{x} = 0 \quad (\text{Hey, this doesn’t make sense!}) ]
If ( f'(c) = 0 ) and ( f''(c) = 0 ) or ( f''(c) ) doesn’t exist, the test is inconclusive.
Wrapping Up
Congratulations, math whiz! You’ve learned how to use the Second Derivative Test to identify those sneaky extrema! Remember, if a function has only one critical point in a domain and it serves as a local max or min, it’s also the global extremum for that interval. 📈🗻
Ready to ace that AP exam? You’ve got this!