From an Average Change to an Instantaneous Rate
An applied model often describes how an output changes as an input varies. The input might be time, production quantity, or another measured variable. Over a whole interval, the model gives an average rate of change. At a particular input, its derivative gives an instantaneous rate of change.
The Mean Value Theorem connects these two descriptions. Provided the model is continuous across the closed interval and differentiable in its interior, there is at least one interior input where the instantaneous rate equals the average rate across the interval. In an application, this identifies a moment or operating level at which the local rate matches the overall rate.
The Mean Value Theorem in a Model
Let \(f\) represent an output as a function of an input \(x\). For distinct inputs \(a\) and \(b\), the average rate of change from \(a\) to \(b\) is the change in output divided by the change in input:
The Mean Value Theorem says that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is at least one \(c\in(a,b)\) such that
The condition \(a<b\) ensures that the denominator is nonzero. The continuity and differentiability conditions matter: without them, the theorem’s existence conclusion is not guaranteed. The point \(c\) is strictly inside the interval; it is not either endpoint.
In an applied setting, \(f'(c)\) is the instantaneous rate at input \(c\). The right-hand side is the average rate over the full interval. Once the endpoint values determine that average, the practical task is to solve the equation \(f'(x)=\frac{f(b)-f(a)}{b-a}\), then retain solutions strictly between \(a\) and \(b\).
Why the Rate-Matching Equation Holds
The theorem’s equation is more than a useful rule for solving examples: it expresses precisely how the change across an interval relates to a derivative somewhere inside it. The relation can be written in either of two equivalent forms:
The second form says that the total change in the model equals the rate at the special input \(c\), multiplied by the interval length. The value of \(c\) is guaranteed by the Mean Value Theorem when its conditions hold; it is not generally known in advance.
Confirm continuity on \([a,b]\) and differentiability on \((a,b)\).
Use the two endpoint outputs and divide their difference by \(b-a\).
Solve \(f'(x)=\frac{f(b)-f(a)}{b-a}\).
Verify that each reported input lies in \((a,b)\) and satisfies the rate equation.
Worked Examples
Worked Example: A Temperature Model
Suppose a temperature model, in degrees, is \(T(t)=20+3t+\frac{1}{2}t^2\), where \(t\) is time in hours. Find a time between \(t=0\) and \(t=4\) when the instantaneous rate of temperature change equals the average rate over the interval.
The model is a polynomial, so it is continuous on \([0,4]\) and differentiable on \((0,4)\). The Mean Value Theorem applies. First calculate the endpoint values:
The average rate over the four-hour interval is
Differentiate the model and set the instantaneous rate equal to \(5\):
Since \(0<2<4\), this is an interior time. Checking the rate gives \(T'(2)=3+2=5\), which matches the average rate. Thus, at \(t=2\) hours, the instantaneous temperature-change rate equals the average rate of \(5\) degrees per hour.
Worked Example: Marginal and Average Cost Change
Let a cost model be \(C(q)=100+4q+q^2\), where \(q\) is the production quantity. Find a production quantity between \(q=2\) and \(q=8\) where the instantaneous rate of cost change equals the average rate of cost change across that range.
Because \(C\) is a polynomial, it is continuous on \([2,8]\) and differentiable on \((2,8)\). Evaluate the endpoints:
The average rate of cost change is
The instantaneous rate is \(C'(q)=4+2q\). Set it equal to the average rate and solve:
The value \(5\) lies strictly between \(2\) and \(8\). Substitution checks the rate: \(C'(5)=4+2(5)=14\). Therefore, at a production quantity of \(5\), the instantaneous cost-change rate matches the average cost-change rate across the interval.
Worked Example: More Than One Matching Time
Consider the model \(M(t)=100+t+\sin(2\pi t)\) on \(0\leq t\leq1\). Find every time in the open interval when its instantaneous rate matches its average rate from \(t=0\) to \(t=1\).
The model is continuous on \([0,1]\) and differentiable on \((0,1)\), so the Mean Value Theorem applies. Since \(\sin(0)=\sin(2\pi)=0\), the endpoint values and average rate are
Differentiate and equate the instantaneous rate to \(1\):
Subtract \(1\) from both sides and divide by \(2\pi\), which is nonzero:
For \(0<t<1\), the solutions are \(t=\frac14\) and \(t=\frac34\). At either time, \(2\pi t\) is respectively \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\), and the cosine is \(0\). Thus \(M'(t)=1\) at both inputs. This example shows that the theorem guarantees at least one matching input, not exactly one.
Existence Is Not the Same as Uniqueness
The Mean Value Theorem guarantees a matching interior input if the conditions are satisfied. It does not, by itself, identify that input, and it does not guarantee that only one such input exists. The examples show two possibilities: a model can have one matching input, or it can have several.
For some models, solving the rate equation exactly is not straightforward. The theorem still establishes that a solution exists, even if the equation cannot be solved in a convenient exact form. In that case, further methods may be needed to locate or approximate a value. Any proposed value must still be checked against the interval and the rate equation.
The theorem also does not say that the instantaneous rate at every input equals the average rate. It identifies at least one interior input where they agree. Nor does it say the matching point is the midpoint. That happens for the quadratic model in the first example, but the periodic model has two matching times.
A Common Mistake: Using Total Change as a Rate
A frequent error is to use \(f(b)-f(a)\), the total change in output, as though it were the average rate. The average rate also divides by the change in input. For the temperature model, the incorrect calculation might be:
This calculation sets the instantaneous rate equal to the total temperature change, rather than to the average rate. It also produces \(t=17\), which is not inside the interval \((0,4)\). The correction is to divide the total change by the interval length:
The corrected value lies in the interval and gives \(T'(2)=5\). Always distinguish the total change \(f(b)-f(a)\) from the average rate \(\frac{f(b)-f(a)}{b-a}\).
Why Rate Matching Matters
In an applied model, the average rate summarizes an interval, while a derivative describes local behavior at one input. The Mean Value Theorem ensures that these descriptions coincide somewhere inside the interval, provided the model has the required continuity and differentiability. This can give a concrete interpretation of an otherwise abstract theorem: at some moment, a process is changing at exactly its overall average pace.
When using this conclusion, keep the logic in order. First verify the theorem’s assumptions. Then compute the average rate from endpoint values. Finally solve the derivative equation and check the candidate inputs. A calculation of \(f'(c)\) without the conditions does not establish that a suitable \(c\) must exist; the theorem’s hypotheses are what justify the existence claim.
Check Your Understanding
Use the rate-matching equation and the theorem’s conditions to answer these questions.
- For \(f(x)=x^2+2x\) on \([1,4]\), calculate the average rate and find an interior input where the instantaneous rate matches it.
- What continuity and differentiability conditions are required to apply the Mean Value Theorem on \([a,b]\)?
- Why is \(\frac{f(b)-f(a)}{b-a}\) an average rate, whereas \(f(b)-f(a)\) is a total change?
- Does the Mean Value Theorem guarantee exactly one matching input? Explain.
- A proposed matching input is an endpoint of the interval. Does that satisfy the theorem’s conclusion? Explain.
