1. First Derivative Test: Pick an $x$ to the left and right of the critical point. Check the sign of $f'$.
- $+ o -$ is a Max. $- o +$ is a Min.
2. Second Derivative Test: Find $f''$ and plug in your critical $x$.
- Positive $(+)$ = Concave UP = Minimum.
- Negative $(-)$ = Concave DOWN = Maximum.
Using $f(x) = x^2 - 4x + 1$, the critical point is at $x = 2$.
Test $x=1$ (Left Neighbor): What is the sign of $f'(1)$?
Test $x=3$ (Right Neighbor): What is the sign of $f'(3)$?
Classify: Based on the neighbor test, is $x=2$ a Maximum or a Minimum?
Find $f'(x)$ and $f''(x)$, then classify the critical points.
$f(x) = -x^2 + 10x - 5$
1. Solve $f'(x)=0$ to find critical $x$.
2. Use $f''(x)$ to classify it.
$f(x) = 2x^3 - 6x$
1. Find the two critical points.
2. Use $f''(x) = 12x$ to test each one.
If $f''(x) = 0$ at a critical point, the Second Derivative Test "fails." Why do you think this happens? Can a curve be perfectly flat and NOT be a peak or a valley? (Hint: Think about $y = x^3$ at $x=0$).
_________________________________________________________________________
_________________________________________________________________________
Find the inflection point ($f''=0$) for $f(x) = x^3 - 6x^2 + 12x$.
An arch for a bridge follows the equation $y = rac{-1}{10}x^2 + x$.
1. Find the maximum height of the arch.
2. Use the Second Derivative test to prove it is a maximum.
Objective: Explain Concavity using faces to a younger student.
The Activity:
1. Draw a smiley face. Point to the bottom of the mouth. "This is a stable bowl. It's a happy Minimum."
2. Draw a frowny face. Point to the top of the mouth. "This is a shaky hill. It's a sad Maximum."
The Lesson: "God wants our hearts to be 'Smiley Faces'—Concave Up—so we can hold His love at our lowest points."
Response: ___________________________________________________________