The Intermediate Value Theorem is a powerful tool in calculus, guaranteeing that continuous functions take on all values between their endpoints. It's like a bridge connecting the dots of a function, ensuring no gaps or jumps exist.

This theorem helps us prove the existence of solutions to equations and approximate roots. It's super useful in real-world applications, from finding equilibrium prices in economics to predicting object collisions in physics.

The Intermediate Value Theorem

Intermediate Value Theorem statement

• States if a function $f$ is continuous on a closed interval $[a, b]$ and $f(a) \neq f(b)$, then for any value $y$ between $f(a)$ and $f(b)$, there exists a value $c$ in the open interval $(a, b)$ such that $f(c) = y$
• Intuitively means a continuous function takes on all values between its endpoints without any gaps or jumps
• Applies to real-valued functions that are continuous on a closed interval
• Examples:
  • If a continuous function has values of -1 and 5 at the endpoints of $[0, 2]$, it must also take on the values 0, 1, 2, 3, and 4 somewhere within the interval
  • The function $f(x) = x^3 - x$ on the interval $[-1, 1]$ satisfies the conditions of the IVT

Proving equation solutions

• To prove an equation $f(x) = y$ has a solution using the IVT:
  • Identify an interval $[a, b]$ where the function $f$ is continuous
  • Verify that $f(a)$ and $f(b)$ lie on opposite sides of the target value $y$
    • Either $f(a) < y < f(b)$ or $f(b) < y < f(a)$
  • The IVT guarantees the existence of a value $c \in (a, b)$ such that $f(c) = y$, proving the equation has at least one solution
• Examples:
  • To show $x^3 - x - 1 = 0$ has a solution between 1 and 2, note $f(1) = -1 < 0$ and $f(2) = 5 > 0$
  • Proving $\sin(x) = 0.5$ has a solution on $[0, \pi]$ since $\sin(0) = 0 < 0.5 < 1 = \sin(\pi/2)$

Root approximation with IVT

• To approximate a root of a continuous function $f$ on an interval $[a, b]$:
  • Confirm $f(a)$ and $f(b)$ have opposite signs, ensuring the existence of a root by the IVT
  • Bisect the interval at the midpoint $m = \frac{a + b}{2}$ and evaluate $f(m)$
    • If $f(m) = 0$, then $m$ is the exact root
    • If $f(m)$ has the same sign as $f(a)$, the root lies in the right half-interval $[m, b]$
    • If $f(m)$ has the same sign as $f(b)$, the root lies in the left half-interval $[a, m]$
  • Iteratively repeat the bisection process with the new, smaller interval until the desired precision is achieved
• This method is known as the bisection method or binary search
• Examples:
  • Approximating a root of $f(x) = x^2 - 2$ on $[1, 2]$ since $f(1) < 0$ and $f(2) > 0$
  • Finding a solution to $\cos(x) = x$ on $[0, 1]$ as $\cos(0) > 0$ and $\cos(1) < 1$

Applications of Intermediate Value Theorem

• The IVT has numerous real-world applications in fields such as physics, engineering, and economics, where continuous functions model various phenomena
  • Proving the existence of solutions to equations describing physical systems
    • Demonstrating the existence of a time when a projectile reaches a specific height
    • Showing two objects must collide or meet at some point in time
  • Approximating zeros or roots of continuous functions representing physical quantities
    • Finding the equilibrium price that balances supply and demand in a market
    • Locating the neutral buoyancy depth of a submarine
• Steps to solve application problems using the IVT:
  1. Recognize the continuous function $f$ and the target value $y$ in the context of the problem
  2. Establish a closed interval $[a, b]$ where $f$ is continuous and $f(a)$ and $f(b)$ fall on opposite sides of $y$
  3. Conclude the existence of a solution or approximate the solution using the bisection method
• Examples:
  • A ball thrown upward must reach a height of 5 meters at some point if its initial height is 0 meters and its peak height is 10 meters
  • The temperature in a room must equal the thermostat setting at some point if the room starts cooler than the setting and later becomes warmer than the setting