Applications of Derivatives

Part 1 of 7 — Critical Points & Increasing/Decreasing

Critical Points

A critical point of $f$ occurs where:

• $f'(x) = 0$, or
• $f'(x)$ is undefined (but $f(x)$ exists)

First Derivative Test for Increasing/Decreasing

Worked Example

Find where $f(x) = x^3 - 3x + 1$ is increasing and decreasing.

$f'(x) = 3x^2 - 3 = 3(x+1)(x-1)$

Critical points: $x = -1$ and $x = 1$.

Critical Points 🎯

First Derivative Test for Local Extrema

At a critical point $c$:

Example (continued)

For $f(x) = x^3 - 3x + 1$:

• At $x = -1$: $f'$ changes from $+$ to $-$ → local max at $f(-1) = 3$
• At $x = 1$: $f'$ changes from $-$ to $+$ → local min at $f(1) = -1$

Classify Critical Points 🎯

Key Takeaways — Part 1

1. Critical points occur where $f' = 0$ or $f'$ is undefined
2. $f' > 0$ → increasing, $f' < 0$ → decreasing
3. First Derivative Test: sign change determines max/min/neither

Applications of Derivatives

Part 2 of 7 — Second Derivative & Concavity

Concavity

Inflection Points

An inflection point is where concavity changes. This occurs where $f'' = 0$ or $f''$ is undefined, AND $f''$ actually changes sign.

Second Derivative Test

At a critical point where $f'(c) = 0$:

• If $f''(c) > 0$: local minimum (concave up)
• If $f''(c) < 0$: local maximum (concave down)
• If $f''(c) = 0$: inconclusive (use First Derivative Test)

Worked Example

$f(x) = x^3 - 6x^2 + 9x + 1$

$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. Critical points: $x = 1, 3$.

$f''(x) = 6x - 12$.

• $f''(1) = -6 < 0$ → local max at $x = 1$
• $f''(3) = 6 > 0$ → local min at $x = 3$
• $f''(x) = 0$ at $x = 2$ → inflection point

Second Derivative Analysis 🎯

Key Takeaways — Part 2

1. $f'' > 0$: concave up; $f'' < 0$: concave down
2. Inflection points: where $f''$ changes sign
3. Second Derivative Test: faster than First Derivative Test when $f''(c) \neq 0$

Applications of Derivatives

Part 3 of 7 — Absolute (Global) Extrema

Extreme Value Theorem (EVT)

If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains an absolute maximum and absolute minimum on $[a, b]$.

Candidates Test (Closed Interval Method)

1. Find all critical points in $(a, b)$
2. Evaluate $f$ at each critical point AND at the endpoints $a$ and $b$
3. The largest value is the absolute max; the smallest is the absolute min

Worked Example

Find the absolute extrema of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.

$f'(x) = 3x^2 - 3 = 0$ → $x = \pm 1$ (both in the interval).

Absolute max = 3 (at $x = -1$ and $x = 2$). Absolute min = $-1$ (at $x = -2$ and $x = 1$).

Absolute Extrema 🎯

Key Takeaways — Part 3

1. EVT guarantees extrema on closed intervals for continuous functions
2. Candidates Test: evaluate at critical points AND endpoints
3. Compare all values to find the absolute max and min

Applications of Derivatives

Part 4 of 7 — Curve Sketching

Complete Curve Sketching Procedure

1. Domain of $f$
2. Intercepts: set $y = 0$ (x-intercepts) and $x = 0$ (y-intercept)
3. Symmetry: even ($f(-x) = f(x)$) or odd ($f(-x) = -f(x)$)
4. First derivative: critical points, increasing/decreasing, local extrema
5. Second derivative: concavity, inflection points
6. End behavior: $\lim_{x \to \pm\infty} f(x)$
7. Asymptotes (if any)

Quick Example

$f(x) = x^4 - 4x^3$

• $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$

  • Critical points: $x = 0, 3$
  • Decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$
  • Local min at $x = 3$: $f(3) = 81 - 108 = -27$

• $f''(x) = 12x^2 - 24x = 12x(x-2)$

  • Inflection at $x = 0$ and $x = 2$
  • Concave up on $(-\infty, 0)$ and $(2, \infty)$
  • Concave down on $(0, 2)$

Curve Sketching from Derivatives 🎯

Given $f'(x) = (x-1)^2(x-4)$:

Key Takeaways — Part 4

1. A systematic approach using both $f'$ and $f''$ gives a complete picture
2. From a graph of $f'$, you can deduce everything about $f$'s shape
3. This is a frequent AP free-response topic

Applications of Derivatives

Part 5 of 7 — Mean Value Theorem

Statement (MVT)

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c$ in $(a,b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Geometric meaning: There's a point where the tangent line is parallel to the secant line through $(a, f(a))$ and $(b, f(b))$.

Worked Example

$f(x) = x^3$ on $[0, 2]$.

Average rate of change: $\frac{f(2)-f(0)}{2-0} = \frac{8-0}{2} = 4$.

Find $c$: $f'(c) = 3c^2 = 4$ → $c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155$.

Since $c \in (0, 2)$, MVT is confirmed.

Mean Value Theorem 🎯

Key Takeaways — Part 5

1. MVT: there's a point where instantaneous rate = average rate
2. Must check: $f$ continuous on $[a,b]$, differentiable on $(a,b)$
3. MVT is used heavily in AP justifications and proofs

Applications of Derivatives

Part 6 of 7 — Related Rates (Mini-Review)

Related Rates Strategy

1. Draw a picture and label variables
2. Write an equation relating the variables
3. Differentiate both sides with respect to time $t$
4. Substitute known values and solve

Related Rates & Applications 🎯

Key Takeaways — Part 6

1. Related rates: differentiate with respect to $t$ using Chain Rule
2. Always identify what you're given and what you're finding
3. Don't substitute values until AFTER differentiating

Applications of Derivatives — Review

Part 7 of 7 — Comprehensive Assessment

Final Assessment 🎯

Applications of Derivatives — Complete! ✅

You have mastered:

• ✅ Critical points and increasing/decreasing analysis
• ✅ Second derivative test and concavity
• ✅ Absolute extrema on closed intervals
• ✅ Curve sketching
• ✅ Mean Value Theorem
• ✅ Related rates applications