From a Bound on Rate to a Bound on Total Change
A derivative describes a function’s rate of change at an individual point. Often, however, the information we have is not the exact derivative at every point, but a range containing all its possible values. The Mean Value Theorem lets us turn that range into a bound on the function’s total change between two endpoints.
This is useful when we know how fast a quantity can change but do not know its exact values throughout an interval. A lower derivative bound limits how far the function can decrease, while an upper derivative bound limits how far it can increase. Together they confine the endpoint increment to an explicit interval.
The Increment Bound
Suppose the derivative stays between two constants throughout the interval. The interval must be one on which \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), so that the Mean Value Theorem applies.
Because \(a<b\), the interval length \(b-a\) is positive. The derivative bounds therefore retain their order when multiplied by that length. The result is a bound on the signed increment, not necessarily on the amount by which the function rises or falls at every point.
If \(f(a)\) is known, adding it to all three parts of the inequality gives bounds on the other endpoint value:
The bounds are inclusive: equality is possible. For example, if \(f'(x)=m\) throughout the interval, then the increment is exactly \(m(b-a)\). The same holds for the upper bound when \(f'(x)=M\) throughout.
Why the Bound Holds
The Mean Value Theorem states that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is some \(c\in(a,b)\) such that the secant slope equals the derivative at \(c\). In symbols, \(f(b)-f(a)=f'(c)(b-a)\). Since the derivative bounds hold at every interior point, they hold at this particular \(c\) as well.
There is a point \(c\in(a,b)\) with \(f(b)-f(a)=f'(c)(b-a)\).
At that point, \(m\leq f'(c)\leq M\).
Since \(b-a>0\), multiplying preserves the inequalities and gives \(m(b-a)\leq f'(c)(b-a)\leq M(b-a)\).
Substitute \(f'(c)(b-a)=f(b)-f(a)\) to obtain the stated bound.
This argument is short, but it captures the central idea: the total change is the derivative at some intermediate point multiplied by the interval length. Bounding that derivative therefore bounds the total change.
Worked Examples
Worked Example: Bounding an Increasing Function’s Increment
Let \(f(x)=x^2+2x\) on \([1,2]\). Bound \(f(2)-f(1)\) using derivative bounds, then check the actual increment.
First, \(f\) is a polynomial, so it is continuous on \([1,2]\) and differentiable on \((1,2)\). Its derivative is
For \(1<x<2\), we have \(2<2x<4\), so \(4<f'(x)<6\). In particular, \(4\leq f'(x)\leq6\) throughout the open interval. The interval length is \(2-1=1\), so the increment theorem gives
Now calculate the endpoint values: \(f(2)=2^2+2(2)=4+4=8\), and \(f(1)=1^2+2(1)=1+2=3\). Thus \(f(2)-f(1)=8-3=5\), which satisfies \(4\leq5\leq6\). The derivative bounds successfully confine the total change.
Worked Example: Using an Initial Value to Bound a Later Value
Define \(F(x)=7+2(x-2)+\frac14(x-2)^2\) on \([2,5]\). Use the derivative bounds \(1\leq F'(x)\leq4\) to bound \(F(5)\), given \(F(2)=7\), and check the result.
Differentiate the formula:
For \(2<x<5\), the quantity \(x-2\) lies between \(0\) and \(3\). Therefore \(2<F'(x)<2+\frac12(3)=\frac72\), which implies \(1\leq F'(x)\leq4\). The interval length is \(5-2=3\), so
Add \(7\) throughout to obtain \(10\leq F(5)\leq19\). Check the actual endpoint value directly:
Since \(\frac{61}{4}=15.25\), the result is within the bound: \(10\leq15.25\leq19\). Equivalently, the actual increment is \(\frac{61}{4}-7=\frac{33}{4}=8.25\), and \(3\leq8.25\leq12\).
Worked Example: Bounding a Decrease
Let \(q(x)=10-3x\) on \([1,4]\). Use the derivative bounds \(-5\leq q'(x)\leq-2\) to bound the increment and verify it exactly.
The function is linear, hence continuous on \([1,4]\) and differentiable on \((1,4)\). Its derivative is \(q'(x)=-3\), which satisfies \(-5\leq-3\leq-2\). The interval length is \(4-1=3\). Applying the theorem gives
Calculate \(q(4)=10-3(4)=10-12=-2\) and \(q(1)=10-3(1)=10-3=7\). Thus \(q(4)-q(1)=-2-7=-9\), and indeed \(-15\leq-9\leq-6\). A negative increment is consistent with both derivative bounds being negative.
Reading the Bounds Carefully
The values \(m\) and \(M\) need not both be positive. If \(m<0<M\), the theorem may bound the increment on both sides of zero, without determining whether the endpoint value is larger or smaller than the starting value. For example, if \(-2\leq f'(x)\leq3\) on an interval of length \(4\), then \(-8\leq f(b)-f(a)\leq12\). This says that the net change cannot be below \(-8\) or above \(12\); it does not select one direction.
A familiar special case is a bound on the absolute derivative. If \(|f'(x)|\leq K\) throughout the interval, then \(-K\leq f'(x)\leq K\). The increment theorem consequently gives
This is an absolute bound on the increment: \(|f(b)-f(a)|\leq K(b-a)\). It still depends on the same continuity and differentiability conditions required by the Mean Value Theorem.
A Common Mistake: Forgetting the Interval Length
A derivative bound controls change per unit of input, not total change by itself. The interval length must be included. For instance, take \(g(x)=3x\) on \([0,3]\). Here \(g'(x)=3\), so \(2\leq g'(x)\leq4\). An incorrect calculation might claim \(2\leq g(3)-g(0)\leq4\), simply copying the derivative bounds. But \(g(3)=9\), \(g(0)=0\), and the increment is \(9\), so that claim is false.
The correct calculation uses \(b-a=3\):
The correction is to multiply both derivative bounds by the positive interval length before interpreting them as bounds on the increment.
Choosing and Applying a Bound
In an application, derivative bounds might be given directly, or they might need to be established from a formula for \(f'\). After obtaining them, keep the order of the argument clear:
- Identify the endpoints \(a<b\) and check continuity on \([a,b]\) and differentiability on \((a,b)\).
- Find constants \(m\) and \(M\) that satisfy \(m\leq f'(x)\leq M\) at every interior point.
- Multiply by the positive length \(b-a\) to bound \(f(b)-f(a)\).
- If one endpoint value is known, add or subtract it to obtain a bound on the other endpoint value.
- Check the signs and, when the function is explicit, verify the conclusion using its endpoint values.
Bounds can be informative even when they are not exact. A wider interval for the derivative leads to a wider interval for the increment; a narrower derivative bound leads to a tighter increment bound. The theorem does not say that every number between the two increment bounds is attained. It guarantees only that the actual increment lies within them.
The essential connection is simple: a pointwise rate bound, valid across the interval, constrains the net change across that interval. The Mean Value Theorem supplies the precise link between the two.
Check Your Understanding
Use the increment bound carefully in each question. Assume the continuity and differentiability hypotheses hold wherever needed.
- If \(a<b\), \(4\leq f'(x)\leq7\) on \((a,b)\), and \(b-a=2\), what bounds follow for \(f(b)-f(a)\)?
- If \(f(a)=5\), \(a<b\), \(b-a=3\), and \(-2\leq f'(x)\leq1\), what interval must contain \(f(b)\)?
- Why does multiplying the derivative bounds by \(b-a\) preserve their order in the theorem?
- If \(|f'(x)|\leq6\) on an interval of length \(2\), what absolute bound follows for the endpoint increment?
- What does the increment theorem conclude when the derivative bounds satisfy \(m<0<M\)?
