Math 231E: Lecture 11 - Mean Value Theorem and Its Applications
Mean Value Theorem (MVT)
Theorem 1 (Mean Value Theorem):
Conditions:
Let f(x) be a function.
f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).
Conclusion:
There exists a number c such that a < c < b.
At this point c, the instantaneous rate of change is equal to the average rate of change over the interval [a, b].
This is expressed by the formula: f'(c) = \frac{f(b) - f(a)}{b - a}.
Corollaries of the Mean Value Theorem
Rolle's Theorem
Theorem 2 (Rolle's Theorem):
Conditions:
Let f(x) be a function.
f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).
The function values at the endpoints are equal: f(a) = f(b).
Conclusion:
There exists a number c such that a < c < b.
At this point c, the instantaneous rate of change is zero: f'(c) = 0.
Example Applying MVT
Example 1: If a car travels 2 miles in one minute, then at some instant, its speed must have been exactly 120 mph.
This is an application of the Mean Value Theorem, where the average speed over the minute is 120 mph (2 miles / 1 minute = 120 miles / hour), implying there was an instant when the instantaneous speed was also 120 mph.
Functions with Zero Derivative
Theorem 3: If f'(x) = 0 for all x \in (a, b), then f(x) is constant on (a, b).
Proof:
Choose any two distinct points x1, x2 \in (a, b).
Without loss of generality, assume x1 < x2.
By the Mean Value Theorem, applied to the interval [x1, x2], there exists some c \in (x1, x2) such that:
f(x2) - f(x1) = f'(c)(x2 - x1)
Given that f'(x) = 0 for all x \in (a, b), it follows that f'(c) = 0.
Substituting this into the MVT equation:
f(x2) - f(x1) = 0 \cdot (x2 - x1) = 0
Therefore, f(x2) = f(x1).
Since x1 and x2 were arbitrary points in (a, b), this means the function's value is the same at any two points, proving f(x) is constant on (a, b).
Functions with Identical Derivatives
Theorem 4: If f(x) and g(x) are two functions defined on (a, b) such that their derivatives are equal, i.e., f'(x) = g'(x) for all x \in (a, b), then f(x) and g(x) differ by a constant.
This means f(x) = g(x) + C for some constant C.
Proof:
Define a new function h(x) = f(x) - g(x).
Differentiate h(x):
h'(x) = f'(x) - g'(x)
Since we are given f'(x) = g'(x), it follows that:
h'(x) = 0
By Theorem 3, if the derivative of a function is zero everywhere in an interval, then the function itself is constant on that interval.
Therefore, h(x) = C for some constant C.
Substituting back the definition of h(x):
f(x) - g(x) = C
Rearranging gives: f(x) = g(x) + C.
Importance of "For All Points in the Interval"
Example 2: The previous theorems rely on the conditions holding true for every point in the specified interval.
Consider the function f(x) = |x|/x on the interval (-1, 1).
This function can be written as:
f(x) = 1 for x \in (0, 1).
f(x) = -1 for x \in (-1, 0).
f(x) is undefined at x=0.
Its derivative, f'(x), is 0 for all x \in (-1, 1) except at x=0. At x=0, the derivative is undefined (the function itself is discontinuous, not differentiable, and undefined at this point).
Clearly, f(x) = |x|/x is not a constant function on (-1, 1).
This example illustrates that even if the derivative is zero at "almost every" point, if there's just one point where the condition fails (e.g., the function is not differentiable or the derivative is not zero), the conclusion of Theorem 3 (that the function is constant) does not hold.
Relationship Between Derivative Sign and Function Behavior
Theorem 5 (Increasing/Decreasing Test):
If f'(x) < 0 on (a, b), then f(x) is decreasing on (a, b).
If f'(x) > 0 on (a, b), then f(x) is increasing on (a, b).
Proof (for f'(x) < 0):
Assume f'(x) < 0 for all x \in (a, b).
Consider any two points x1, x2 \in (a, b) such that x1 < x2.
By the Mean Value Theorem, applied to the interval [x1, x2], there exists a number c such that x1 < c < x2 where:
f(x2) - f(x1) = f'(c)(x2 - x1)
We know that f'(c) < 0 (because c \in (a, b) and we assumed f'(x) < 0 for all x \in (a, b)).
We also know that x2 - x1 > 0 (since x1 < x2).
Therefore, the product f'(c)(x2 - x1) must be negative:
f'(c)(x2 - x1) < 0
This implies that f(x2) - f(x1) < 0.
Rearranging, f(x2) < f(x1).
Since for any x1 < x2 in the interval, f(x2) < f(x1), this proves that f(x) is decreasing on (a, b).
Proof (for f'(x) > 0): (Similar logic, omitted in transcript but implied)
If f'(x) > 0 and x2 - x1 > 0, then f'(c)(x2 - x1) > 0. So, f(x2) - f(x1) > 0, which means f(x2) > f(x1), indicating an increasing function.