5.3 Curve Analysis & The First Derivative Test

5.3 Curve Analysis & The First Derivative Test

Core CLEP Concept: Curve analysis is the cornerstone of differential calculus application. By investigating the behavior of the first derivative $f'(x)$, we can rigorously deduce where a function is rising or falling, locate its local and global extreme values, and reconstruct the exact geometric trajectory of $y = f(x)$ without plotting hundreds of arbitrary points.

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1. Critical Numbers and Critical Points

Before finding the maximum or minimum values of any function, we must identify the candidate points where such extrema could possibly occur.

Formal Definition of a Critical Number

Critical Number: A number $c$ is defined as a critical number of a function $f$ if and only if:

1. $c$ is in the domain of $f$ ($f(c)$ exists and is a defined real number), AND
2. Either $f'(c) = 0$ (a stationary point) OR $f'(c)$ is undefined / does not exist (a singular point).

If $c$ is a critical number, the coordinate pair $(c, f(c))$ is called a critical point on the graph of $f$.

The Domain Requirement (A Major CLEP Trap)

A value $x = c$ where $f'(c)$ does not exist is NOT a critical number if $f(c)$ itself is undefined:

• For $f(x) = \frac{1}{x}$, $f'(x) = -\frac{1}{x^2}$. Both $f(0)$ and $f'(0)$ are undefined. However, $x = 0$ is a vertical asymptote, NOT a critical number, because $0$ is not in the domain of $f$.
• For $f(x) = x^{2/3}$, $f(0) = 0$ is defined, but $f'(x) = \frac{2}{3x^{1/3}}$ is undefined at $x = 0$. Therefore, $x = 0$ is a valid critical number (corresponding to a sharp cusp).

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2. Increasing and Decreasing Test (Monotonicity)

The sign of the first derivative over an open interval dictates the direction of the function's curve.

The Monotonicity Theorem

Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$:

1. If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$ (the curve rises from left to right).
2. If $f'(x) < 0$ for all $x \in (a, b)$, then $f$ is strictly decreasing on $[a, b]$ (the curve falls from left to right).
3. If $f'(x) = 0$ for all $x \in (a, b)$, then $f$ is constant on $[a, b]$.

5-Step Sign Chart Protocol for $f'(x)$

1. Establish Domain: Identify all domain restrictions (asymptotes, holes, radical constraints).
2. Differentiate and Factor: Compute $f'(x)$ and write it as a single factored algebraic fraction.
3. Identify Partition Numbers: Set numerator $= 0$ (stationary points) and denominator $= 0$ (singular points / discontinuities).
4. Partition the Number Line: Divide the real domain into disjoint open intervals using the partition numbers.
5. Test Sample Points: Choose a convenient test number $k$ inside each interval and evaluate the sign of $f'(k)$ ($+$ or $-$).

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3. The First Derivative Test for Local Extrema

Fermat's Theorem guarantees that if $f$ has a local extremum at $c$, then $c$ must be a critical number. However, not every critical number yields an extremum. We classify critical points using the First Derivative Test.

   LOCAL MAXIMUM (+ to -)             LOCAL MINIMUM (- to +)            NO EXTREMUM (+ to +)
      f'(x) > 0   f'(x) < 0              f'(x) < 0   f'(x) > 0             f'(x) > 0   f'(x) > 0
         / \                                \     /                           /       /
        /   \                                \   /                           /   .--'
       /  *  \                                \ * /                         /   /  *
          c (Local Max)                         c (Local Min)                     c (Plateau)

Formal Statement of the First Derivative Test

The First Derivative Test: Let $c$ be a critical number of a continuous function $f$:

1. Local Maximum: If $f'(x)$ changes from positive to negative ($+ \to -$) as $x$ increases through $c$, then $f(c)$ is a local maximum.
2. Local Minimum: If $f'(x)$ changes from negative to positive ($- \to +$) as $x$ increases through $c$, then $f(c)$ is a local minimum.
3. No Local Extremum: If $f'(x)$ does not change sign at $c$ (remains $+ \to +$ or $- \to -$), then $f(c)$ is neither a local maximum nor a local minimum.

Step-by-Step Worked Example: Full Polynomial Curve Analysis

Problem: Find all intervals of increase/decrease and all local extrema for $f(x) = 2x^3 - 9x^2 + 12x - 3$

1. Domain: All real numbers $\mathbb{R}$.
2. Differentiate and factor: $$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$
3. Critical numbers: Set $f'(x) = 0 \implies x = 1$ and $x = 2$.
4. Construct the Sign Chart:

1. Classify Local Extrema:
   • At $x = 1$: $f'$ changes from $+$ to $-$ $\implies$ Local Maximum at $(1, f(1)) = (1, 2(1)^3 - 9(1)^2 + 12(1) - 3) = (1, 2)$.
   • At $x = 2$: $f'$ changes from $-$ to $+$ $\implies$ Local Minimum at $(2, f(2)) = (2, 2(8) - 9(4) + 12(2) - 3) = (2, 1)$.

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4. Absolute (Global) Extrema & The Closed Interval Method

While local extrema describe peaks and valleys relative to neighboring points, absolute (global) extrema represent the highest and lowest values across an entire domain.

The Extreme Value Theorem (EVT)

If $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ is guaranteed to attain an absolute maximum value and an absolute minimum value at least once on $[a, b]$.

The Closed Interval Method (The Candidates Test)

To find the absolute maximum and minimum of a continuous function $f$ on $[a, b]$:

1. Find all critical numbers of $f$ that lie strictly inside the open interval $(a, b)$.
2. Evaluate $f(c)$ at each interior critical number $c$.
3. Evaluate $f(a)$ and $f(b)$ at the endpoints.
4. Compare values: The greatest functional value is the absolute maximum; the least is the absolute minimum.

Step-by-Step Worked Example: Closed Interval Candidates Test

Problem: Find the absolute maximum and minimum values of $f(x) = x^4 - 4x^3 + 2$ on the closed interval $[-1, 4]$.

1. Differentiate and find critical numbers: $$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) = 0 \implies x = 0, \quad x = 3$$ Both critical numbers $0$ and $3$ lie within $(-1, 4)$.
2. Evaluate $f(x)$ at all candidate points:
   • Left Endpoint: $f(-1) = (-1)^4 - 4(-1)^3 + 2 = 1 + 4 + 2 = 7$
   • Critical Point 1: $f(0) = (0)^4 - 4(0)^3 + 2 = 2$
   • Critical Point 2: $f(3) = (3)^4 - 4(3)^3 + 2 = 81 - 108 + 2 = -25$
   • Right Endpoint: $f(4) = (4)^4 - 4(4)^3 + 2 = 256 - 256 + 2 = 2$
3. Conclusion:
   • Absolute Maximum Value: $7$, occurring at $x = -1$.
   • Absolute Minimum Value: $-25$, occurring at $x = 3$.

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5. Decoding the Graph of the Derivative $f'(x)$

A staple CLEP Calculus question presents candidates with the graph of the derivative $y = f'(x)$ and asks for properties of the original function $y = f(x)$.