Unit 5: Function Analysis Using Derivatives

Determining Intervals of Increase and Decrease

To understand the behavior of a function $f(x)$ without graphing it, we look to the first derivative, $f'(x)$. The first derivative represents the instantaneous rate of change or the slope of the tangent line.

Critical Numbers

Before identifying intervals, you must find the Critical Numbers (or Critical Points). A critical number of a function $f$ is a number $c$ in the domain of $f$ such that:

1. f'(c) = 0
2. f'(c) \text{ is undefined}

These are the only locations where a function can change direction (from increasing to decreasing or vice versa).

The Test for Increasing/Decreasing Functions

Once you have identified critical numbers, you divide the domain into test intervals. Within each interval, determine the sign of $f'(x)$.

• Increasing: If f'(x) > 0 for all $x$ in an interval $(a, b)$, then $f$ is increasing on $[a, b]$.
• Decreasing: If f'(x) < 0 for all $x$ in an interval $(a, b)$, then $f$ is decreasing on $[a, b]$.
• Constant: If f'(x) = 0 for all $x$ in an interval, then $f$ is constant.

Worked Example

Problem: Find the open intervals on which $f(x) = x^3 - rac{3}{2}x^2$ is increasing or decreasing.

Solution:

1. Find the derivative:
   f'(x) = 3x^2 - 3x

2. Find Critical Numbers: Set $f'(x) = 0$.
   3x(x - 1) = 0
   x = 0, \; x = 1

3. Create a Sign Chart: Test values between the critical numbers.

Conclusion: $f$ is increasing on $(-\infty, 0)$ and $(1, \infty)$ and decreasing on $(0, 1)$.

First Derivative Test

The First Derivative Test is used to determine if a critical number corresponds to a Relative (Local) Maximum, a Relative Minimum, or neither.

The Theorem

Let $c$ be a critical number of a continuous function $f$.

1. Relative Maximum: If $f'(x)$ changes from positive to negative at $c$, then $f$ has a local maximum at $(c, f(c))$.
2. Relative Minimum: If $f'(x)$ changes from negative to positive at $c$, then $f$ has a local minimum at $(c, f(c))$.
3. Neither: If $f'(x)$ is positive on both sides of $c$ or negative on both sides of $c$, then $f(c)$ is neither a max nor a min (it might be a terrace point).

Justification on the AP Exam

When asked to justify a relative extremum, avoid vague phrases like "the slope changes" or "it goes up then down." be specific:

"Function $f$ has a relative maximum at $x=c$ because $f'(x)$ changes from positive to negative at this point."

Determines Concavity and Points of Inflection

While the first derivative tells us direction (increasing/decreasing), the second derivative $f''(x)$ tells us the curvature or shape of the graph.

Testing for Concavity

• Concave Up: The graph lies above its tangent lines. Mathematically, f''(x) > 0 on an interval.
  • Think: The graph is shaped like a cup $\cup$ (holds water).
  • Note: If $f$ is concave up, $f'$ is increasing.
• Concave Down: The graph lies below its tangent lines. Mathematically, f''(x) < 0 on an interval.
  • Think: The graph is shaped like a frown $\cap$ (spills water).
  • Note: If $f$ is concave down, $f'$ is decreasing.

Points of Inflection (POI)

A Point of Inflection happens at a point $(c, f(c))$ where the graph intersects its tangent line and changes concavity.

Condition for POI:

1. $c$ is in the domain of $f$ (assuming $f$ is continuous).
2. f''(c) = 0 OR f''(c) \text{ is undefined}.
3. Crucial: $f''(x)$ must change sign at $x=c$ (from + to - or - to +).

Memory Aid: The $f, f', f''$ Ladder

Understanding the relationship hierarchy is vital:

Second Derivative Test

The Second Derivative Test is an alternative method to classify critical numbers as relative max or min. It uses concavity to determine the nature of a flat spot on the curve.

The Rules

Let $f$ be a function such that $f'(c) = 0$ (meaning we have a horizontal tangent) and $f''$ exists on an open interval containing $c$.

1. Relative Minimum: If f''(c) > 0, then $f$ is concave up at $c$. A horizontal tangent in a "cup" shape must be a bottom.
   • Result: Local Minimum.
2. Relative Maximum: If f''(c) < 0, then $f$ is concave down at $c$. A horizontal tangent in a "frown" shape must be a peak.
   • Result: Local Maximum.
3. Inconclusive: If f''(c) = 0, the test fails. You must go back and use the First Derivative Test.

Worked Example

Problem: Classify the critical points of $f(x) = x^4 - 4x^3$ using the Second Derivative Test.

Solution:

1. Find $f'(x)$ and Critical Numbers:
   f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)
   $f'(x) = 0$ at $x=0$ and $x=3$.

2. Find $f''(x)$:
   f''(x) = 12x^2 - 24x

3. Apply Test:

   • At $x=3$: f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36
     Since $36 > 0$, $f$ is Concave Up. Therefore, $x=3$ is a Relative Minimum.
   • At $x=0$: f''(0) = 12(0)^2 - 24(0) = 0
     Since $f''(0)=0$, the test is inconclusive. We must use the First Derivative Test (Sign Chart). ( Checking chart: $f'$ is negative left of 0 and negative right of 0, so $x=0$ is neither.)

Common Mistakes & Pitfalls

1. "It's a POI because $f''(x)=0$": This is false. $f''(x)=0$ creates a candidate for a Point of Inflection. You must verify that the sign of $f''$ actually changes. For example, $f(x)=x^4$ has $f''(0)=0$, but it is not a POI because it is concave up everywhere.
2. Confusing the Tests: Students often mix up the First and Second Derivative Tests. Remember: The First Deriv Test looks at the sign change of $f'$ around $c$. The Second Deriv Test looks at the sign of $f''$ exactly at $c$.
3. Assuming undefined derivatives aren't extrema: Critical points occur where $f'(x)=0$ AND where $f'(x)$ is undefined (like a sharp corner or cusp defined by $f(x) = |x|$ or $x^{2/3}$). Relative extrema can exist at these sharp points.
4. Endpoint Errors: The First and Second Derivative Tests define Relative (Local) extrema, which occur inside open intervals. Absolute extrema can occur at endpoints, but the derivative tests don't "find" endpoints automatically; you must check them separately in optimization problems.