From a Derivative Bound to a Function-Change Bound
A derivative describes how a function changes locally. The Mean Value Theorem connects that local information to a difference in function values across an entire interval. In particular, if the derivative’s magnitude never exceeds a fixed number, then the function cannot change by more than that number multiplied by the interval’s length.
This conclusion can be useful even when the exact value of a function at one endpoint is difficult to calculate. A known value and a derivative bound can give a guaranteed range for other function values.
The Function-Change Inequality
Here is the precise result. The derivative bound is required at every interior point of the interval between the two values being compared. The function need not be differentiable at the endpoints, but it must be continuous there.
The same conclusion applies to any two points in an interval: if \(u\) and \(v\) lie in an interval where the hypotheses hold, then \[ |f(v)-f(u)|\leq M|v-u|. \] For \(u=v\), both sides are zero. If the points are distinct, apply the theorem to the interval with those points as endpoints.
In addition to bounding the absolute change, the inequality gives two-sided bounds. If \(a<b\), then \[ f(a)-M(b-a)\leq f(b)\leq f(a)+M(b-a). \] These follow from the standard absolute-value inequality \(|z|\leq r\), which implies \(-r\leq z\leq r\), using \(z=f(b)-f(a)\) and \(r=M(b-a)\).
Why the Inequality Follows
The proof uses the Mean Value Theorem. That theorem says that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then some \(c\in(a,b)\) satisfies \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] The hypotheses in the function-change theorem are exactly what is needed to use this statement.
For some \(c\in(a,b)\), \(f(b)-f(a)=f'(c)(b-a)\).
Since \(b-a>0\), \(|f(b)-f(a)|=|f'(c)|(b-a)\).
The point \(c\) is inside the interval, so \(|f'(c)|\leq M\). Therefore \(|f(b)-f(a)|\leq M(b-a)\).
This proves the result. The important connection is that the Mean Value Theorem makes the overall change equal to a derivative value at some interior point multiplied by the interval length. A bound that holds at every interior point therefore bounds that particular derivative value too.
Worked Examples
Worked Example: Bounding the Change in a Polynomial
Let \(f(x)=x^2\) on \([1,3]\). Find a bound for \(|f(3)-f(1)|\) using a bound on the derivative, then check the actual change.
First, \(f\) is a polynomial, so it is continuous on \([1,3]\) and differentiable on \((1,3)\). Its derivative is \[ f'(x)=2x. \] For \(1<x<3\), \(2<2x<6\), so in particular \(|f'(x)|\leq 6\). We can take \(M=6\), and the interval length is \(3-1=2\). The theorem gives \[ |f(3)-f(1)|\leq 6(2)=12. \] To check the actual difference, \(f(3)=3^2=9\) and \(f(1)=1^2=1\). Thus \[ |f(3)-f(1)|=|9-1|=8\leq 12. \] The bound is valid, though it is not an exact value for the change.
Worked Example: Bounding the Change in a Radical Function
Let \(f(x)=\sqrt{x+3}\) on \([1,6]\). Find a bound for the change between the endpoint values and compare it with the actual change.
The quantity \(x+3\) is positive throughout the interval, so the function is continuous there and differentiable in its interior. Differentiating gives \[ f'(x)=\frac{1}{2\sqrt{x+3}}. \] For \(1<x<6\), \(x+3>4\), so \(\sqrt{x+3}>2\). Therefore \[ |f'(x)|=\frac{1}{2\sqrt{x+3}}<\frac14, \] and the weaker bound \(|f'(x)|\leq \frac14\) holds. The interval length is \(6-1=5\), so \[ |f(6)-f(1)|\leq \frac14(5)=\frac54. \] Now calculate the actual values: \(f(6)=\sqrt9=3\) and \(f(1)=\sqrt4=2\). Hence \[ |f(6)-f(1)|=|3-2|=1\leq\frac54. \] The theorem guarantees the bound without needing to use the exact endpoint difference to obtain it.
Worked Example: Bounding the Change in a Rational Function
Let \(f(x)=\frac{1}{x+1}\) on \([0,2]\). Find a derivative bound and use it to estimate the endpoint change.
The denominator is nonzero on \([0,2]\), so \(f\) is continuous on that closed interval and differentiable in its interior. Its derivative is \[ f'(x)=-\frac{1}{(x+1)^2}. \] For \(0<x<2\), \(x+1>1\), which gives \[ |f'(x)|=\frac{1}{(x+1)^2}<1. \] Thus \(|f'(x)|\leq 1\), so we may take \(M=1\). The interval length is \(2-0=2\), and the theorem gives \[ |f(2)-f(0)|\leq 1(2)=2. \] Direct evaluation gives \(f(2)=\frac13\) and \(f(0)=1\). Therefore \[ |f(2)-f(0)|=\left|\frac13-1\right| =\left|-\frac23\right|=\frac23\leq 2. \] The calculated difference confirms the estimate.
Choosing and Using a Derivative Bound
The number \(M\) does not have to be the smallest possible bound. Any nonnegative number that satisfies \(|f'(x)|\leq M\) at every required interior point is valid. A smaller valid \(M\) produces a tighter estimate, because the resulting bound \(M(b-a)\) is smaller.
It is the absolute value of the derivative that matters. A derivative may be negative, which means the function is decreasing, but a large negative derivative can still correspond to a large change in magnitude. For example, if \(f'(x)=-4\) throughout an interval of length \(2\), then \(|f'(x)|=4\), and the theorem gives a change bound of \(4(2)=8\), not a bound based on the negative sign.
The conclusion can also be applied from a known value at one point. If \(|f'(x)|\leq M\) between \(a\) and a point \(x\), then \[ |f(x)-f(a)|\leq M|x-a|. \] Equivalently, \[ f(a)-M|x-a|\leq f(x)\leq f(a)+M|x-a|. \] This gives an interval that must contain \(f(x)\). It may be wider than necessary, but it is guaranteed by the assumptions.
Common Pitfall: Checking Only the Endpoints
A bound on the derivative must hold throughout the interval, not merely at its endpoints. Checking two endpoint values of \(f'\) does not in general establish what happens between them.
For a concrete example, let \(g(x)=x^2(1-x)^2\) on \([0,1]\). Differentiating gives \[ g'(x)=2x(1-x)(1-2x). \] At both endpoints, \(g'(0)=0\) and \(g'(1)=0\). If those endpoint values were mistakenly treated as proof that \(g'\) is zero throughout the interval, one might set \(M=0\). That would predict no change between \(0\) and \(\frac12\). But \[ g(0)=0,\qquad g\left(\frac12\right)=\left(\frac12\right)^2\left(1-\frac12\right)^2 =\frac14\cdot\frac14=\frac1{16}. \] So the actual change is \(\frac1{16}\), not zero. The problem is that the proposed derivative bound \(M=0\) does not hold in the interior. A valid application requires a bound across the entire interval between the points being compared.
Another common error is to use an upper bound on \(f'(x)\) as if it were a bound on \(|f'(x)|\). For instance, \(f'(x)=-4\) satisfies \(f'(x)\leq 1\), but \(|f'(x)|=4\), not at most \(1\). To bound the size of a function change using this theorem, check the absolute value.
Why the Result Is Useful
The inequality turns information about a derivative into a guarantee about function values. It can show that two values must be close when their inputs are close, provided the derivative stays bounded. It can also provide a range for an unknown value when a nearby value is known. These are estimates rather than claims that the function changes at a constant rate: the theorem only guarantees the stated upper bound.
Every application rests on three checks:
- The function is continuous on the closed interval between the points.
- The function is differentiable on the open interval between the points.
- A single nonnegative \(M\) bounds \(|f'(x)|\) at every point in that open interval.
Once these conditions are established, the conclusion is direct: the absolute difference in function values is at most \(M\) times the interval length.
Check Your Understanding
For each question, identify the derivative bound and the interval length when they are needed.
- If \(|f'(x)|\leq 3\) between \(x=2\) and \(x=5\), what bound does the theorem give for \(|f(5)-f(2)|\), assuming its continuity and differentiability hypotheses hold?
- Why does the theorem require a bound on \(|f'(x)|\), rather than only an upper bound on \(f'(x)\)?
- For \(f(x)=x^2+1\) on \([0,2]\), find a valid bound for \(|f(2)-f(0)|\) using the derivative, then calculate the actual difference to check it.
- Do values of \(f'\) at just the two endpoints establish a bound for \(|f'|\) throughout an interval? Explain.
- If \(|f'(x)|\leq M\) between \(a\) and \(x\), write the two-sided estimate for \(f(x)\) in terms of \(f(a)\), \(M\), and \(|x-a|\).
