#Extreme Value Theorem Study Notes

#Table of Contents

1. What is the Extreme Value Theorem?
2. Exam Tip
3. Worked Example
4. Practice Questions
5. Glossary
6. Summary and Key Takeaways

#Extreme Value Theorem

#What is the Extreme Value Theorem?

The extreme value theorem states that:

1. If a function $f$ is continuous over the closed interval $[a, b]$, then:

   • $f$ has at least one minimum value and at least one maximum value on the interval $[a, b]$.

2. These values might occur at the endpoints of the interval. If a minimum or maximum value does not occur at an endpoint, it will be a local minimum or local maximum within the open interval.

   • In this case, if the function is differentiable on $(a, b)$, then the derivative of the function will be zero at that point.

#Exam Tip

#Worked Example

A social sciences researcher uses a function $m(t)$ to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood. The function $m(t)$ is twice-differentiable, with $t$ measured in days.

The table below gives selected values of $m(t)$ over the time interval $0 \le t \le 12$:

Question: Justify why there must be at least one time, $t$, for $7 \le t \le 12$, at which $m'(t)$, the rate of change of the mass, equals 0 kilograms per day.

Answer: To prove that the derivative has a particular value at an unspecified point, we can use the extreme value theorem:

1. Continuity: Start by showing that $m(t)$ is continuous. Since $m(t)$ is twice-differentiable, it implies that $m(t)$ is differentiable, which in turn implies that $m(t)$ is continuous on $[7, 12]$.

2. Application of Extreme Value Theorem: By the extreme value theorem, $m(t)$ must have a maximum value on $[7, 12]$.

3. Endpoints: Check the endpoints:

   • $m(7) = 70.3 < 89.7 = m(10)$
   • $m(10) = 89.7 > 89.1 = m(12)$

   Hence, the maximum value does not occur at either of the endpoints of the interval.

4. Local Maximum: Because $m(t)$ is differentiable on $(7, 12)$, the maximum must be a local maximum point somewhere in that open interval, at which point $m'(t)$ equals zero.

Thus, there exists at least one time $t$ in the interval $7 \le t \le 12$ where the rate of change of the mass, $m'(t)$, is zero.

#Practice Questions

Practice Question

Q1: Given a continuous function $f(x)$ on the interval $[1, 5]$, identify the points where $f(x)$ might attain its minimum and maximum values.

Practice Question

Q2: Explain why a differentiable function on an open interval $(a, b)$ implies that the function is continuous on $(a, b)$.

Practice Question

Q3: Consider a function $g(x)$ that is continuous on $[2, 8]$ and differentiable on $(2, 8)$. If $g(2) = 3$, $g(5) = 7$, and $g(8) = 3$, determine whether $g(x)$ has a local minimum or maximum in the interval $(2, 8)$.

#Glossary

• Continuous Function: A function with no breaks, jumps, or holes in its graph.
• Differentiable Function: A function that has a derivative at all points in its domain.
• Local Maximum/Minimum: The highest or lowest value of a function in a small interval around a point.
• Endpoint: The values at the ends of a closed interval.

#Summary and Key Takeaways

• The extreme value theorem guarantees that a continuous function on a closed interval $[a, b]$ has both a maximum and minimum value.
• These extreme values may occur at the endpoints or within the interval.
• If the function is also differentiable, any local maximum or minimum within the interval will have a derivative of zero.
• Always verify the conditions of continuity and differentiability when applying the theorem.