Theorem Applications

Theorem Applications

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

Topic Overview

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)

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 $5$ is between $2$ and $7$, by the Intermediate Value Theorem there exists $c \in (a,b)$ such that $f(c) = 5$."

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 $[0,1]$: $f(1) = 1 - 4 + 1 = -2 < 0$.

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

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

Theorem Applications

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

Statement

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

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

Geometric Meaning

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

MVT vs. IVT

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

Worked Example

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

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

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

AP Writing Template

"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

Theorem Applications

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

Statement

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

The Closed Interval Method

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

When Does EVT Fail?

Worked Example

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 $x = \pm 1$.

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!

Theorem Applications

Part 4 of 7 — Rolle’s Theorem

Statement

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

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

Rolle’s Theorem vs. MVT

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

Geometric Meaning

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

Worked Example

$f(x) = x^2 - 4x + 3$ 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

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 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

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

Theorem Applications

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

AP FRQ Theorem Patterns

Complete Worked FRQ

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

(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

Theorem Applications

Part 7 of 7 — Comprehensive Assessment

Theorem Reference

Common AP Mistakes

Assessment — Set 1 🎯

Assessment — Set 2 🎯

Identify the theorem. 🔍

Final challenge. ✍️

Theorem Applications — Complete! 🎓

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