Welcome! In this section, we will explore the Mean Value Theorem (MVT), an important result that connects a function's average rate of change over an interval to its instantaneous rate of change at a specific point. For example, if your average speed on a trip is 60 mph, the MVT guarantees that at some moment, your speedometer must have read exactly 60 mph. This theorem forms a crucial theoretical bridge for many other concepts we will learn.
Before tackling the Mean Value Theorem in its full generality, we'll look at a simpler, special case named after the mathematician Michel Rolle. Rolle's Theorem deals with functions that start and end at the same height.
Let $f$ be a function that satisfies three conditions:
Then, there is a number $c$ in $(a, b)$ such that $f'(c) = 0$.
What does this mean visually? If a smooth curve starts and ends at the same y-value, it must have at least one point in between where the tangent line is horizontal. It might go up and then come back down, or down and then back up. In either case, it must "turn around," and at that turning point, the derivative is zero. In the graph below, click on the circles next to the function names to see three different examples that all satisfy the conditions on their respective intervals.
Verify that the function $f(x) = x^3 - 4x$ satisfies the hypotheses of Rolle's Theorem on the interval $[0, 2]$, and find all numbers $c$ that satisfy the conclusion.
Step 1: Check the conditions.
Since all three conditions are satisfied, Rolle's Theorem guarantees there's at least one $c$ in $(0, 2)$ where $f'(c)=0$.
Step 2: Find the value(s) of $c$.
First, find the derivative: $f'(x) = 3x^2 - 4$.
Now, set it to zero and solve for $x$:
$$3x^2 - 4 = 0 \implies 3x^2 = 4 \implies x^2 = \frac{4}{3} \implies x = \pm\sqrt{\frac{4}{3}} = \pm\frac{2}{\sqrt{3}}$$
Step 3: Check if $c$ is in the interval.
We have two potential values, but we only care about the one inside the open interval $(0, 2)$. The value $c = -2/\sqrt{3}$ is outside the interval. The value $c = 2/\sqrt{3} \approx 1.155$ is inside $(0, 2)$.
Conclusion: The value $c = 2/\sqrt{3}$ satisfies the conclusion of Rolle's Theorem for $f(x)$ on $[0, 2]$.
The Mean Value Theorem is a "tilted" version of Rolle's Theorem. Instead of requiring the function to start and end at the same height ($f(a)=f(b)$), we now allow the endpoints to be different. The MVT states that there is a point where the instantaneous rate of change (slope of the tangent) equals the average rate of change over the whole interval (slope of the secant line connecting the endpoints).
Let $f$ be a function that satisfies two conditions:
Then, there is a number $c$ in $(a, b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$The expression on the right, $\frac{f(b) - f(a)}{b - a}$, is the slope of the secant line connecting the points $(a, f(a))$ and $(b, f(b))$. The term on the left, $f'(c)$, is the slope of the tangent line at the point $c$. The MVT guarantees that there is at least one place $c$ where the tangent line is parallel to the secant line.
In the interactive graph below, the red line is the secant line. Move the slider for 'a' to find the point where the tangent line has the same slope as the secant line.
Consider $f(x) = x^3 - x$ on the interval $[0, 2]$. Find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
Step 1: Check conditions. $f(x)$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. The MVT applies.
Step 2: Calculate the slope of the secant line.
First find the function values at the endpoints:
$f(0) = 0^3 - 0 = 0$
$f(2) = 2^3 - 2 = 8 - 2 = 6$
Now calculate the slope:
$$m_{secant} = \frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3$$
Step 3: Find the derivative and set it equal to the secant slope.
$f'(x) = 3x^2 - 1$.
We want to find $c$ such that $f'(c) = 3$:
$$3c^2 - 1 = 3 \implies 3c^2 = 4 \implies c^2 = \frac{4}{3} \implies c = \pm\frac{2}{\sqrt{3}}$$
Step 4: Select the value(s) in the interval.
The interval is $(0, 2)$. The value $c = -2/\sqrt{3}$ is not in this interval. The value $c = 2/\sqrt{3} \approx 1.155$ is in the interval.
Conclusion: The value that satisfies the MVT is $c = 2/\sqrt{3}$.
A car travels from a starting point and after 5 minutes its distance from the start is 4 miles. Does the Mean Value Theorem guarantee there was a time when the car's instantaneous speed was exactly 48 mph?
The MVT is powerful not just for its direct application, but for the important results we can prove with it. Two of its most important consequences relate to functions with zero derivatives and functions that share the same derivative.
If $f'(x) = 0$ for all $x$ in an open interval $(a, b)$, then $f$ is a constant function on $(a, b)$.
This seems obvious, but the MVT provides the rigorous proof. If a function weren't constant, you could find two points $x_1$ and $x_2$ with different y-values ($f(x_1) \neq f(x_2)$). The secant line between them would have a non-zero slope. By the MVT, the derivative at some point $c$ between them would have to equal this non-zero slope. This contradicts our starting fact that the derivative is always zero.
If $f'(x) = g'(x)$ for all $x$ in an open interval $(a, b)$, then $f(x) - g(x) = C$ for some constant $C$. That is, $f(x) = g(x) + C$.
This is incredibly important! It tells us that if two functions have the same derivative, they can only differ by a constant. They are just vertical shifts of one another. This is the entire reason we add "+ C" when we find an antiderivative (or indefinite integral).
Proof of the Corollary: This result is a clever application of the previous theorem.
1. Define a new function: Let $h(x) = f(x) - g(x)$.
2. Differentiate it: Since $f$ and $g$ are differentiable, so is $h$. The derivative is $h'(x) = f'(x) - g'(x)$.
3. Use the given information: We are told that $f'(x) = g'(x)$. This means that $h'(x) = f'(x) - f'(x) = 0$ for all $x$ in the interval.
4. Apply the previous theorem: Since $h'(x)=0$ everywhere on the interval, the theorem tells us that $h(x)$ must be a constant. Let's call it $C$.
5. Conclude: If $h(x) = C$, then by our definition, $f(x) - g(x) = C$. This proves the corollary.
Can you apply the Mean Value Theorem to $f(x) = |x-2|$ on the interval $[1, 4]$? Why or why not?
Suppose $f$ is a differentiable function such that $f(1)=10$ and $f'(x) \ge 2$ for all $x$ in the interval $[1, 4]$. What is the smallest possible value of $f(4)$?
Show that the equation $f(x) = x^3 + x - 1 = 0$ has exactly one real root.
Today we explored the Mean Value Theorem, one of the cornerstones of calculus. We started with the special case of Rolle's Theorem and then generalized its conclusion. The main takeaway is the connection it builds between average and instantaneous rates of change.
After this lecture, you should be able to: