From a Derivative Bound to a Function-Change Bound
A derivative describes how rapidly a function is changing at an individual input. A bound on the derivative throughout an interval gives more: it limits how much the function can change between any two inputs in that interval.
The key is that the Mean Value Theorem connects a change across an interval to the derivative at some point inside it. If the derivative’s absolute value is never greater than \(M\), then the change in function value cannot exceed \(M\) times the distance between the inputs.
The estimate is useful because it turns local information about derivatives into a guarantee about a finite change. The bound applies to every eligible pair of inputs, not just to a pair selected in advance.
Precise Statement and Conditions
Let \(I\) be an interval, let \(f\) be a function defined on \(I\), and let \(M\geq 0\). Suppose that for every pair \(a<b\) in \(I\), \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \(|f'(t)|\leq M\) for every \(t\in(a,b)\). Then the bound below holds for every \(x,y\in I\).
If \(I\) is an open interval and \(f\) is differentiable throughout \(I\), the continuity condition on each closed segment follows from differentiability there. The statement also allows endpoints of the segment to be points where a derivative is not defined: differentiability is required only in the open segment, while continuity is required at its endpoints.
The absolute values in both parts of the estimate matter. The quantity \(|f(y)-f(x)|\) measures the size of the function-value change without regard to whether the function rises or falls. The quantity \(|y-x|\) is the distance between the inputs.
Why the Estimate Follows
First consider two distinct inputs and label the smaller one \(a\) and the larger one \(b\). The Mean Value Theorem says that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), there is a point \(c\in(a,b)\) such that the slope between the endpoints equals the derivative at \(c\). The derivative bound can then be applied at that point.
For \(a<b\), the hypotheses give some \(c\in(a,b)\) such that \(f(b)-f(a)=f'(c)(b-a)\).
Since \(b-a>0\), \(|f(b)-f(a)|=|f'(c)|\,|b-a|\).
Because \(|f'(c)|\leq M\), multiplication by \(|b-a|\geq0\) gives \(|f(b)-f(a)|\leq M|b-a|\).
This proves the result when the inputs are ordered. For arbitrary distinct \(x,y\), set \(a=\min\{x,y\}\) and \(b=\max\{x,y\}\). The absolute differences are unchanged by swapping the inputs, so the same inequality holds. If \(x=y\), both sides are zero. Thus the estimate holds for every pair.
The proof uses the derivative bound at the particular point \(c\) guaranteed by the Mean Value Theorem. It does not require finding that point. It does require the bound to hold throughout the open segment, because the theorem does not tell us in advance where \(c\) will be.
Applying the Bound
Worked Example: A Quadratic on a Closed Interval
Let \(f(t)=t^2\) on \([0,2]\). Bound the change between the inputs \(x=\frac12\) and \(y=2\) using a uniform absolute derivative bound.
Differentiate:
For every \(t\in(0,2)\), \(0<2t<4\), so \(|f'(t)|\leq4\). We may use \(M=4\). The function is continuous on the interval and differentiable in its interior, so the estimate applies.
The input distance is
Therefore, the change in function values is bounded by
Check the actual change directly: \(f(2)=2^2=4\), and \(f(\frac12)=(\frac12)^2=\frac14\). Hence
Since \(\frac{15}{4}\leq6\), the calculated values satisfy the bound. This particular estimate is not exact: it guarantees a change no larger than \(6\), while the actual change is \(\frac{15}{4}\).
Worked Example: A Trigonometric Function
Let \(f(t)=\sin t\) on \([0,\frac{\pi}{3}]\). Use the derivative bound to control the change from \(0\) to \(\frac{\pi}{3}\).
Differentiate and bound the derivative:
Thus \(M=1\) works throughout the interval. The function is continuous on the closed interval and differentiable in its interior. The distance between the inputs is \(\frac{\pi}{3}-0=\frac{\pi}{3}\), so
Checking the endpoint values gives \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) and \(\sin(0)=0\). The actual change is therefore \(\frac{\sqrt{3}}{2}\). It is indeed no greater than \(\frac{\pi}{3}\): \(\sqrt{3}<2\) gives \(\frac{\sqrt{3}}{2}<1\), while \(\pi>3\) gives \(1<\frac{\pi}{3}\). Hence \(\frac{\sqrt{3}}{2}<\frac{\pi}{3}\), as required.
When the Bound Is Exact
The estimate can be sharp: there are functions and input pairs for which the bound equals the actual change. Consequently, no smaller universal constant can replace \(M\) in the conclusion when all we know is that \(|f'|\leq M\).
Worked Example: A Sharp Bound for a Linear Function
Let \(f(t)=-3t+4\), and compare the inputs \(-2\) and \(1\). Differentiating gives
Thus we can take \(M=3\). The input distance is
The estimate gives
Now calculate the endpoint values: \(f(1)=-3(1)+4=1\), and \(f(-2)=-3(-2)+4=6+4=10\). Therefore, \(|f(1)-f(-2)|=|1-10|=9\). The actual change equals the bound, so the estimate is exact in this example.
A Common Misinterpretation
A bound on the absolute value of the derivative is stronger than a one-sided statement that the derivative is at most some number. The latter does not, by itself, limit the size of a negative derivative. Therefore it cannot be substituted for the hypothesis \(|f'(t)|\leq M\) when the goal is to bound the absolute change in function values.
Worked Example: Why a One-Sided Bound Is Not Enough
Consider \(f(t)=-10t\) on \([0,1]\). Its derivative is \(f'(t)=-10\), so \(f'(t)\leq1\) throughout the interval. A mistaken argument might treat this as though \(|f'(t)|\leq1\), and conclude that the change over an input distance of \(1\) is at most \(1\).
But the absolute derivative is \(|f'(t)|=|-10|=10\), not \(1\). Direct calculation confirms the problem: \(f(1)=-10\), \(f(0)=0\), and \(|f(1)-f(0)|=|-10-0|=10\). The proposed bound \(10\leq1\) is false.
The correction is to use the actual absolute derivative bound \(M=10\). The estimate then gives \(|f(1)-f(0)|\leq10\cdot|1-0|=10\), which matches the computed change.
Another error is checking the derivative bound at only a few points and assuming it holds between them. The theorem requires a bound throughout the open segment. A finite collection of derivative values alone does not establish that condition. First justify a uniform bound on the whole interval; only then apply the estimate.
Interpreting the Result
The conclusion is a guaranteed maximum size for the change, not a prediction of the exact change. The bound may be conservative, as in the quadratic and sine examples, or exact, as in the linear example. A larger valid choice of \(M\) still gives a correct inequality, but usually a less informative one.
For a useful estimate, choose a bound that is as small as you can justify across the entire interval. The derivative need not equal \(M\) anywhere; \(M\) only needs to be an upper bound for \(|f'|\) at every interior point. Once that uniform control is established, the Mean Value Theorem turns it into control over every pair of function values.
The logical pattern is concise: verify continuity and differentiability on the segment, establish \(|f'|\leq M\) throughout its interior, and multiply \(M\) by the input distance. Each condition has a role, and omitting the absolute value or the interval-wide derivative bound can invalidate the conclusion.
Check Your Understanding
For each question, identify the derivative bound and the interval conditions that justify the estimate.
- If \(|f'(t)|\leq5\) throughout a segment and the input distance is \(\frac{2}{3}\), what upper bound follows for the absolute function-value change?
- Why does the proof require a derivative bound throughout the open segment rather than only at its endpoints?
- For \(f(t)=4t+1\), what is the smallest \(M\) that works on every interval, and when is the estimate an equality?
- What is the difference between \(f'(t)\leq M\) and \(|f'(t)|\leq M\) when bounding an absolute function-value change?
- In the quadratic example, why is the resulting bound valid even though it is larger than the actual change?
