🎯⭐ INTERACTIVE LESSON

Theorem Applications

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Theorem Applications - Complete Interactive Lesson

Part 1: The Intermediate Value Theorem (IVT)

Theorem Applications

Part 1 of 7 — The Intermediate Value Theorem (IVT)

Topic Overview

Part	Topic
1	Intermediate Value Theorem (IVT)
2	Mean Value Theorem (MVT)
3	Extreme Value Theorem (EVT)
4	Rolle’s Theorem
5	FTC & Theorem Selection
6	AP-Style Free-Response Workshop
7	Comprehensive Assessment

Statement of the IVT

$\boxed{\text{If } f \text{ is continuous on } [a,b] \text{ and } N \text{ is between } f(a) \text{ and } f(b), \text{ then } \exists\, c \in (a,b): f(c) = N}$

What IVT Tells You (and Doesn’t)

IVT Guarantees	IVT Does NOT Tell You
At least one $c$ exists	Where $c$ is located
$f(c) = N$ for that $c$	How many solutions there are
is in the open interval

Key Fact: IVT requires ONLY continuity. Differentiability is not needed.

AP Writing Template

"Since $f$ is continuous on $[a,b]$ , $f(a) = 2$ and $f(b) = 7$ , and is between and , by the Intermediate Value Theorem there exists such that ."

Worked Example — Proving a Root Exists

Show that $f(x) = x^3 - 4x + 1$ has a root on $[0, 2]$ .

$f(0) = 1 > 0$ and $f(2) = 8 - 8 + 1 = 1 > 0$ . Hmm, both positive. Try : .

Since $f$ is continuous (polynomial), $f(0) = 1 > 0$ and $f(1) = -2 < 0$ . By IVT, such that .

Practice — IVT 🎯

Apply IVT step by step. 🔍

$f$ is continuous. $f(2) = -1$ , $f(6) = 5$ .

Use IVT. ✍️

Key Takeaways — Part 1

IVT requires only continuity on $[a,b]$
Guarantees existence, not location or uniqueness
Target value $N$ must be between $f(a)$ and $f(b)$
Always cite the theorem by name on the AP exam

Part 2: The Mean Value Theorem (MVT)

Theorem Applications

Part 2 of 7 — The Mean Value Theorem (MVT)

Statement

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ :

Part 3: The Extreme Value Theorem (EVT)

Theorem Applications

Part 3 of 7 — The Extreme Value Theorem (EVT)

Statement

$\boxed{\text{If } f \text{ is continuous on } [a,b], \text{ then } f \text{ attains an absolute max and min on } [a,b]}$

Part 4: Rolle's Theorem & MVT Applications

Theorem Applications

Part 4 of 7 — Rolle’s Theorem

Statement

If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f(a) = f(b)$ :

Part 5: FTC and When to Use Each Theorem

Theorem Applications

Part 5 of 7 — FTC & Theorem Selection

The Fundamental Theorem of Calculus

Part 1 (Derivative of an Integral):

$\boxed{\frac{d}{dx}\int_a^x f(t)\,dt = f(x)}$

Part 6: Practice Workshop

Theorem Applications

Part 6 of 7 — AP-Style Free-Response Workshop

AP FRQ Theorem Patterns

Part	Typical Prompt	Theorem
(a)	"Must $f(c) = k$ for some $c$ ?"	IVT
(b)	"Must $f'(c) = m$ for some ?"

Part 7: Final Assessment

Theorem Applications

Part 7 of 7 — Comprehensive Assessment

Theorem Reference

Theorem	Hypothesis	Conclusion
IVT	$f$ continuous on $[a,b]$	$\exists\, c: f(c) = N$ (for between )

f(1) = 1 - 4 + 1 = -2 < 0

\exists\, c \in (0,1)

\exists c \in (a, b) such that f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}

There’s a point where the tangent line is parallel to the secant line connecting $(a, f(a))$ and $(b, f(b))$ .

Feature	IVT	MVT
Hypothesis	Continuous	Continuous + differentiable
Conclusion	$f(c) = N$	$f'(c) = \text{avg. rate}$
Guarantees about	Function values	Derivative values

Key Fact: MVT requires TWO hypotheses: continuity on $[a,b]$ AND differentiability on $(a,b)$ . You must verify both on the AP exam.

$f(x) = x^2$ on $[1, 3]$ . Find the $c$ guaranteed by MVT.

Average rate: $\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4$ .

$f'(c) = 2c = 4 \Rightarrow c = 2 \in (1,3)$ . ✓

"Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , by the Mean Value Theorem there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a} = \ldots$ "

Practice — MVT 🎯

Apply MVT step by step. 🔍

$f(x) = \sqrt{x}$ on $[1, 9]$ .

Find $c$ . ✍️

Key Takeaways — Part 2

MVT: instantaneous rate = average rate at some point
Requires continuity on $[a,b]$ AND differentiability on $(a,b)$
Geometric meaning: tangent parallel to secant
Always verify both hypotheses on the AP exam

The Closed Interval Method

Step	Action
1	Find all critical points: $f'(x) = 0$ or $f'(x)$ DNE
2	Evaluate $f$ at each critical point in $(a,b)$
3	Evaluate $f$ at the endpoints $a$ and $b$
4	The largest value is the absolute max, smallest is the absolute min

Key Fact: Absolute extrema on a closed interval can occur at critical points OR at endpoints. You must check ALL candidates.

Situation	Problem
Open interval $(a,b)$	No guaranteed max/min
$f$ is not continuous	May have a gap; extrema may not exist
$f$ on $(-\infty, \infty)$	Unbounded domain

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

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0$ at $x = \pm 1$ .

$x$	$-2$	$-1$	$1$	$2$
$f(x)$	$-1$	$3$	$-1$	$3$

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

Practice — EVT 🎯

Find absolute extrema step by step. 🔍

$f(x) = 2x^3 - 3x^2$ on $[-1, 2]$ .

Closed interval method. ✍️

Key Takeaways — Part 3

EVT: continuous on $[a,b]$ $\Rightarrow$ absolute max and min exist
Use the closed interval method: check critical points AND endpoints
EVT requires a closed interval and continuity
Extrema can occur at endpoints!

$\boxed{\exists\, c \in (a,b) \text{ such that } f'(c) = 0}$

Rolle’s Theorem vs. MVT

Feature	Rolle’s	MVT
Extra condition	$f(a) = f(b)$	None beyond MVT
Conclusion	$f'(c) = 0$	$f'(c) = \frac{f(b)-f(a)}{b-a}$
Relationship	Special case of MVT	General theorem

Key Fact: Rolle’s Theorem IS the Mean Value Theorem when $f(a) = f(b)$ , because the average rate of change is $\frac{f(b)-f(a)}{b-a} = 0$ .

If the function starts and ends at the same height, it must have a horizontal tangent somewhere in between.

$f(x) = x^2 - 4x + 3$ on $[1, 3]$ . Verify Rolle’s and find $c$ .

Check: $f(1) = 1 - 4 + 3 = 0$ , $f(3) = 9 - 12 + 3 = 0$ . ✓ $f(1) = f(3) = 0$ .

$f$ is a polynomial, so continuous and differentiable everywhere. ✓

$f'(x) = 2x - 4 = 0 \Rightarrow x = 2 \in (1,3)$ . ✓

Real-World Application

If a ball is thrown up and returns to its starting height, at some moment the velocity was exactly zero (at the peak).

Practice — Rolle’s Theorem 🎯

Verify Rolle’s Theorem. 🔍

$f(x) = \sin x$ on $[0, \pi]$ .

Apply Rolle’s Theorem. ✍️

Key Takeaways — Part 4

Rolle’s: $f(a) = f(b)$ + continuous + differentiable $\Rightarrow f'(c) = 0$
Special case of MVT where average rate is $0$
Geometric: horizontal tangent between equal endpoints
Must verify all three conditions

Part 2 (Evaluating Definite Integrals):

$\boxed{\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F' = f}$

With Chain Rule (Variable Upper Limit):

$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$

Theorem Selection Guide

You Want to Show...	Use This Theorem	Key Hypothesis
$f(c) = N$ for some $c$	IVT	Continuity
$f'(c) = m$ for some $c$	MVT	Cont. + diff.
$f'(c) = 0$ for some $c$	Rolle’s	Cont. + diff. + $f(a)=f(b)$
Absolute max/min exist	EVT	Continuity on $[a,b]$
$\frac{d}{dx}\int_a^x f$	FTC Part 1	continuous
$\int_a^b f$ from antiderivative	FTC Part 2	$f$ continuous

AP Tip: On multiple-choice, look for keywords like "must there exist," "guarantee," "show that." These signal a theorem justification.

Practice — Which Theorem? 🎯

Match the scenario. 🔍

FTC Part 1 with chain rule. ✍️

Key Takeaways — Part 5

FTC Part 1: derivative of integral = the integrand
FTC with chain rule: multiply by $g'(x)$
Know which theorem to use based on what you need to prove
IVT for values, MVT for derivatives, EVT for extrema

$f$ is continuous on $[-2, 6]$ and differentiable on $(-2, 6)$ .

$x$	$-2$	$0$	$3$	$6$
$f(x)$	$4$	$1$	$7$	$4$

(a) Must there exist $c \in (-2, 6)$ where $f(c) = 5$ ?

$f(0) = 1 < 5 < 7 = f(3)$ . Since $f$ is continuous on $[0,3]$ , by IVT $\exists\, c \in (0,3) \subset (-2,6)$ with $f(c) = 5$ . ✓

(b) Must there exist $c \in (-2, 6)$ where $f'(c) = 0$ ?

$f(-2) = 4 = f(6)$ . Since $f$ is continuous on $[-2,6]$ and differentiable on $(-2,6)$ , by Rolle’s Theorem $\exists\, c \in (-2,6)$ with $f'(c) = 0$ . ✓

(c) Find the average rate of change on $[-2, 6]$ .

$\frac{f(6) - f(-2)}{6-(-2)} = \frac{4-4}{8} = 0$ .

By MVT, $\exists\, c \in (-2,6)$ with $f'(c) = 0$ . (Same conclusion as Rolle’s — consistent!)

(d) Must $f$ attain an absolute max on $[-2, 6]$ ?

Yes. By EVT, since $f$ is continuous on the closed interval $[-2,6]$ , $f$ attains both an absolute max and an absolute min.

AP-style questions 🎯

$g$ is continuous on $[0,8]$ and differentiable on $(0,8)$ . $g(0) = 2$ , $g(4) = 10$ , $g(8) = 6$ .

Justify with theorems. 🔍

$h$ is continuous and differentiable. $h(1) = 3$ , $h(5) = 3$ , $h(3) = 8$ .

MVT calculation. ✍️

Key Takeaways — Part 6

AP FRQs frequently combine multiple theorems in one problem
Always identify which theorem matches the conclusion needed
Cite theorems by name and verify all hypotheses
IVT for values, MVT/Rolle’s for derivatives, EVT for extrema

Mistake	Correction
Using IVT for derivatives	IVT is for function values; use MVT for derivatives
Not verifying hypotheses	Always check continuity/differentiability
Confusing Rolle’s and MVT	Rolle’s is MVT with $f(a)=f(b)$
Forgetting FTC chain rule	$\frac{d}{dx}\int_a^{g(x)} f = f(g(x))\cdot g'(x)$
Not citing theorem by name	AP requires explicit theorem citation

Assessment — Set 1 🎯

Assessment — Set 2 🎯

Identify the theorem. 🔍

Final challenge. ✍️

Theorem Applications — Complete! 🎓

Part	Topic	Status
1	Intermediate Value Theorem (IVT)	✅
2	Mean Value Theorem (MVT)	✅
3	Extreme Value Theorem (EVT)	✅
4	Rolle’s Theorem	✅
5	FTC & Theorem Selection	✅
6	AP-Style Free-Response Workshop	✅
7	Comprehensive Assessment	✅

You have mastered all the major calculus theorems for the AP exam!

Theorem Applications

Geometric Meaning

MVT vs. IVT

Worked Example

AP Writing Template

Key Takeaways — Part 2

The Closed Interval Method

When Does EVT Fail?

Worked Example

Key Takeaways — Part 3

Rolle’s Theorem vs. MVT

Geometric Meaning

Worked Example

Real-World Application

Key Takeaways — Part 4

With Chain Rule (Variable Upper Limit):

Theorem Selection Guide

Key Takeaways — Part 5

Complete Worked FRQ

Key Takeaways — Part 6

Common AP Mistakes

Theorem Applications — Complete! 🎓