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Proving Existence of Roots and Continuity in Single-Variable Calculus

This document explores key concepts in single-variable calculus, focusing on the proofs of root existence for equations such as cos(θ) = θ, cos(y) = y, cos(x) = x, and √(y) = -1 + y within specified intervals. Utilizing the Intermediate Value Theorem, opposite signs at boundary points confirm the presence of roots. Furthermore, it examines the continuity of functions and demonstrates the limit of x²sin(1/x) as x approaches 0 through the Squeeze Theorem. Essential derivative rules and related rates are also discussed to reinforce understanding.

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Proving Existence of Roots and Continuity in Single-Variable Calculus

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  1. Exam 1 Math 1231: Single-Variable Calculus

  2. Question 1: Limits

  3. Question 2: Implicit Function • 1. Prove there is a root for the equation cosθ=θ in the interval (0, π). • Define f(θ)=cosθ-θ, f(0)=1, f(π)=-1-π, f(θ) have opposite sign on the boundary points, IVP implies there is a root. • 2. Prove there is a root for the equation cosy=y in the interval (0, π). • Define f(y)=cosy-y, f(0)=1, f(π)=-1-π, f(y) have opposite sign on the boundary points, IVP implies there is a root. • 3. Prove there is a root for the equation cosx=x in the interval (0, π). • Define f(x)=cosx-x, f(0)=1, f(π)=-1-π, f(x) have opposite sign on the boundary points, IVP implies there is a root. • Prove there is a root for the equation sqrt(y)=-1+y in the interval (0, 4). • Define f(y)=sqrt(y)+1-y, f(0)=1, f(4)=-1, f(y) have opposite sign on the boundary points, IVP implies there is a root.

  4. Question 2: Implicit Function Use parenthesis!!!

  5. Question 2: Implicit Function

  6. Question 2: Implicit Function

  7. Question 2: Implicit Function

  8. Question 2: Implicit Function

  9. Question 3: Continuity and Derivative • f(x) is continuous only if limx0 f(x)=f(0)=0. • How to show that limx0 x2sin(1/x) =0?

  10. Question 3: Continuity and Derivative Squeeze theorem

  11. Question 3: Continuity and Derivative • f(x) is continuous only if limx0 f(x)=f(0)=0. • How to show that limx0 x2sin(1/x) =0? -1 <= sin(1/x) <= 1  -x2 <= x2sin(1/x) <= x2. Note that -x2 and x2 approach 0 as x goes to 0, so does x2sin(1/x).

  12. Question 3: Continuity and Derivative

  13. Question 4: Derivative Rules

  14. Question 5: Related Rates

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