### Determining Concavity: AP Calculus AB/BC Study Guide

#### Introduction

Hello future calculus wizards! Ready to crack the code of concavity? This magical property of functions tells us if a curve is smiling (concave up) or frowning (concave down). Grab your calculus wand, and let's dive into the enchanted world of second derivatives and inflection points! 🧙♂️✨

#### What is Concavity? 🕵️♀️

In the land of calculus, concavity describes whether a curve bends upwards or downwards. Picture this: if your graph is concave up, it's like a bowl ready to catch rain. If it's concave down, it looks more like the top of a sad umbrella.

Technically, here's what it means:

- A function is concave up when the derivative (slope of the function) is increasing.
- A function is concave down when the derivative is decreasing.

Put simply, if the slopes of the tangents to the graph are getting steeper (in the positive direction), the function is concave up. If they are getting less steep or more negative, the function is concave down.

#### The Magic of the Second Derivative ✨

How do we figure out concavity? 🧙♂️ We call upon the second derivative, which is basically the derivative of the derivative (derivative-ception!). If the first derivative, ( f'(x) ), measures how fast the function is growing or shrinking, the second derivative, ( f''(x) ), tells us how fast ( f'(x) ) itself is changing.

- When ( f''(x) ) > 0, the function is concave up (think of a happy smiley face).
- When ( f''(x) ) < 0, the function is concave down (imagine a sad, frowning face).

#### Inflection Points: The Magical Switches

An inflection point is where the curve switches its concavity, like a plot twist in your favorite fantasy movie. For a point ( x = c ) to be an inflection point, ( f''(c) ) should be zero or undefined, and the second derivative should change sign.

#### Summarizing the Enchantment ✍️

- A graph is concave up if ( f'(x) ) is increasing or ( f''(x) ) > 0.
- A graph is concave down if ( f'(x) ) is decreasing or ( f''(x) ) < 0.
- An inflection point occurs where ( f''(x) = 0 ), and the function changes concavity.

#### Application Time: Work Those Derivatives! 📐

Consider the enchanted function: [ f(x) = x^3 - 3x^2 + 2x + 1 ]

**a) Determine the intervals where ( f(x) ) is concave up and concave down.**
**b) Find where ( f(x) ) has points of inflection.**

**Step-By-Step Spell Cast:**

**Step 1: First Derivative**
[ f'(x) = 3x^2 - 6x + 2 ]

**Step 2: Second Derivative**
[ f''(x) = 6x - 6 ]

**Step 3: Find Possible Inflection Points**
Set ( f''(x) = 0 ):
[ 6x - 6 = 0 ]
[ x = 1 ]

So we have a candidate for an inflection point at ( x = 1 ). This splits the number line into two intervals: ( (-\infty, 1) ) and ( (1, \infty) ).

**Step 4: Test the Intervals**

For ( x = 0 ): [ f''(0) = 6(0) - 6 = -6 ] Since ( f''(0) < 0 ), ( f(x) ) is concave down on ( (-\infty, 1) ).

For ( x = 2 ): [ f''(2) = 6(2) - 6 = 6 ] Since ( f''(2) > 0 ), ( f(x) ) is concave up on ( (1, \infty) ).

✏️ Hence:

- ( f(x) ) is concave down on ( (-\infty, 1) ).
- ( f(x) ) is concave up on ( (1, \infty) ).

**Step 5: Confirm Inflection Points**

We tested ( x = 1 ) and confirmed that concavity changes, making ( x = 1 ) an official point of inflection.

📝 **Concavity Practice Problems**

**Question 1:** Let ( h(x) = 5x^3 ). What is the concavity of ( h ) at ( x=5 )?

**Question 2:** Let ( h(x) = 3x^4 + 2x^3 ). What is the concavity of ( h ) at ( x = -\frac{1}{3} )?

✅ **Concavity Answers and Solutions**

**Question 1:**
[ h'(x) = 15x^2 ]
[ h''(x) = 30x ]
Evaluate at ( x = 5 ):
[ h''(5) = 30 \cdot 5 = 150 ]
Since ( h''(5) > 0 ), ( h ) is concave up at ( x = 5 ).

**Question 2:**
[ h'(x) = 12x^3 + 6x^2 ]
[ h''(x) = 36x^2 + 12x ]
Evaluate at ( x = -\frac{1}{3} ):
[ h''\left( -\frac{1}{3} \right) = 36 \cdot \left( -\frac{1}{3} \right)^2 + 12 \cdot \left( -\frac{1}{3} \right) = 0 ]
Since ( h''\left( -\frac{1}{3} \right) = 0 ), ( x = -\frac{1}{3} ) might be an inflection point. Check for sign changes around ( x = -\frac{1}{3} ):

( -0.4 ) gives positive value, concave up. ( -0.3 ) gives negative value, concave down.

So, ( x = -\frac{1}{3} ) is an inflection point.

#### Conclusion

Bravo! You have unraveled the mysteries of concavity, second derivatives, and inflection points. Whether your graph is smiling or frowning, you now have the mathematical mojo to figure it out. Keep practicing, and you’ll be a concavity conjurer in no time. 🔮✨

Happy Calculus Adventures! 🧠