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Applications of Derivatives

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Applications of Derivatives - Complete Interactive Lesson

Part 1: Critical Points & Increasing/Decreasing

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

$f'(x)$	Behavior of $f$
$f'(x) > 0$

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 $x = 1$ .

Interval	$f'(x)$	Behavior
$(-\infty, -1)$	$+$

Critical Points 🎯

First Derivative Test for Local Extrema

At a critical point $c$ :

Sign change of $f'$	Conclusion
$+ \to -$	Local maximum at

Classify Critical Points 🎯

Key Takeaways — Part 1

Critical points occur where $f' = 0$ or $f'$ is undefined
$f' > 0$ → increasing, → decreasing

Part 2: Second Derivative & Concavity

Applications of Derivatives

Part 2 of 7 — Second Derivative & Concavity

Concavity

$f''(x)$	Concavity
$f''(x) > 0$

Part 3: Absolute (Global) Extrema

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 $[a, b]$ .

Part 4: Curve Sketching

Applications of Derivatives

Part 4 of 7 — Curve Sketching

Complete Curve Sketching Procedure

Domain of $f$
Intercepts: set $y = 0$ (x-intercepts) and $x = 0$ (y-intercept)
Symmetry: even ( $f (- x)$ ) or odd ()

Part 5: Mean Value Theorem

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 such that:

Part 6: Related Rates (Mini-Review)

Applications of Derivatives

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

Related Rates Strategy

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

Related Rates & Applications 🎯

Key Takeaways — Part 6

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

Part 7: Comprehensive Assessment

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

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 $f (1) =$

First Derivative Test: sign change determines max/min/neither

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)

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

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

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

Candidates Test (Closed Interval Method)

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

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).

$x$	$f(x)$
$-2$	$-8+6+1 = -1$
$-1$	$-1+3+1 = 3$
$1$	$1-3+1 = -1$
$2$	$8-6+1 = 3$

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

Absolute Extrema 🎯

Key Takeaways — Part 3

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

First derivative: critical points, increasing/decreasing, local extrema

Second derivative: concavity, inflection points

End behavior:

\lim_{x \to \pm\infty} f(x)

Asymptotes (if any)

$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$
- Critical points: $x = 0, 3$
- Decreasing on $(-\infty, 3)$ , increasing on $(3, \infty)$
- Local min at $x = 3$ :
$f''(x) = 12x^2 - 24x = 12x(x-2)$

Curve Sketching from Derivatives 🎯

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

Key Takeaways — Part 4

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

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

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

$f(x) = x^3$ 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

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

f(3) = 81 - 108 = -27

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

Applications of Derivatives

Example (continued)

Inflection Points

Second Derivative Test

Worked Example

Key Takeaways — Part 2

Candidates Test (Closed Interval Method)

Worked Example

Key Takeaways — Part 3

Quick Example

Key Takeaways — Part 4

Worked Example

Key Takeaways — Part 5