From a Known Value to an Estimate
Suppose a function is known at one input, but its value at a nearby input is difficult to calculate exactly. If we can control how large the derivative is between those inputs, the Mean Value Theorem turns that control into a guaranteed estimate.
The estimate has a simple structure: the unknown value cannot differ from the known value by more than the derivative bound multiplied by the distance between the inputs. The derivative bound must hold throughout the interval, not just at the two endpoints.
Conditions and the Estimate
The conditions are exactly those that allow the Mean Value Theorem to be applied on the interval with endpoints \(a\) and \(x\). In particular, \(a\) and \(x\) must be in the domain, \(f\) must be continuous on the closed interval between them, and differentiable at every point strictly between them. The derivative bound must also hold at every such interior point.
The estimate concerns the magnitude \(|f'(t)|\). A statement that only gives an upper bound \(f'(t)\le M\) is not, by itself, a bound on the magnitude: the derivative could be a large negative number. We will return to this distinction below.
This is an interval estimate for a particular function value, anchored at the reference pair \((a,f(a))\). It does not usually identify the exact value. Instead, it guarantees that the value lies inside a range whose width is determined by the derivative bound and the distance between the inputs.
Why the Estimate Follows from the Mean Value Theorem
Apply the Mean Value Theorem to \(f\) between \(a\) and \(x\). There is some point \(c\) strictly between them such that the change in the function equals the derivative at \(c\) times the change in input.
Taking absolute values and using the assumed bound at \(c\) gives
For any real number \(u\), the inequality \(|u|\le r\), where \(r\ge 0\), is equivalent to \(-r\le u\le r\). Set \(u=f(x)-f(a)\) and \(r=M|x-a|\). Then
Adding \(f(a)\) to all three parts proves the stated estimate. The argument also explains why the distance is an absolute value: the points may occur in either order, but a distance is nonnegative.
Worked Examples
Worked Example: Estimating a Quadratic at a Nearby Input
Let \(f(t)=t^2+1\). Use \(f(3)=10\) to estimate \(f(3.1)\), and find a guaranteed interval for its value.
First, the interval between the inputs is \([3,3.1]\). The function is a polynomial, so it is continuous on this interval and differentiable in its interior. Its derivative is
For \(3\le t\le 3.1\), \(2t\le 6.2\), and \(2t\) is positive. Therefore \(|f'(t)|\le 6.2\) throughout the interval. Here \(M=6.2\), and the input distance is \(|3.1-3|=0.1\). The maximum possible change in the function value under this estimate is
Since the reference value is \(f(3)=3^2+1=10\), the guaranteed range is
Check the estimate by direct evaluation: \(f(3.1)=(3.1)^2+1=9.61+1=10.61\). Indeed, \(9.38\le10.61\le10.62\), as required.
Worked Example: Estimating a Reciprocal from a Reference Value
Let \(f(t)=1/t\). Starting from \(f(2)=1/2\), estimate \(f(11/5)\) using a derivative bound on \([2,11/5]\).
The interval contains only positive inputs, so \(f\) is continuous there and differentiable in its interior. Differentiation gives
Since \(t\ge2\) on the interval, \(t^2\ge4\), and therefore \(|f'(t)|\le1/4\). The input distance is \(11/5-2=1/5\). Thus
Using the reference value \(f(2)=1/2\), we obtain
Direct evaluation gives \(f(11/5)=5/11\). To check that it lies in the estimated range, compare fractions with positive cross-products:
So the exact value lies between the two guaranteed endpoints.
Worked Example: Estimating to the Left of the Reference Input
Let \(f(t)=t^3\). Use \(f(1)=1\) to bound \(f(0.9)\).
The inputs determine the interval \([0.9,1]\). The function is continuous on this interval and differentiable in its interior, with
For \(0.9\le t\le1\), \(0\le3t^2\le3\), so \(|f'(t)|\le3\). The input distance is \(|0.9-1|=0.1\), and the resulting change bound is \(3(0.1)=0.3\). Starting from \(f(1)=1\), we get
The estimate works when the unknown input is to the left of the reference input because the distance is positive. Direct evaluation gives \(f(0.9)=(0.9)^3=0.729\), and \(0.7\le0.729\le1.3\), confirming the bound.
Choosing and Using the Derivative Bound
The constant \(M\) must bound \(|f'|\) over the entire interval between the two inputs. A value that works only at the reference point, or only at the unknown input, is not enough. In the first example, for instance, the derivative changes as \(t\) changes. The value \(6.2\) was chosen because it bounds \(2t\) throughout \([3,3.1]\), not merely because it is the derivative at one endpoint.
A larger valid \(M\) still gives a correct estimate, but the resulting interval will generally be wider. When it is easy to do so, use a smaller valid bound. Do not choose a smaller value just to make the estimate appear more precise: if it fails to bound the derivative somewhere in the interval, the conclusion is not justified.
The theorem also works when \(x<a\). Its formula uses \(|x-a|\), so there is no need to reverse the inputs or change the estimate. The derivative bound and continuity and differentiability conditions must still hold on the whole interval between them.
A Common Mistake: Bounding the Derivative, Not Its Magnitude
A one-sided upper bound \(f'(t)\le M\) does not imply \(|f'(t)|\le M\). For example, let \(f(t)=-100t\) on \([0,1]\), with reference input \(a=0\). Its derivative is \(f'(t)=-100\), so \(f'(t)\le1\) everywhere. But \(f(1)=-100\), and therefore
The incorrect conclusion would be to treat \(f'(t)\le1\) as if it meant \(|f'(t)|\le1\). It does not. To use the two-sided estimate in this tutorial, establish an absolute bound such as \(|f'(t)|\le M\). For the example just given, the valid magnitude bound is \(|f'(t)|=100\), which gives the correct change bound \(100\).
Interpreting the Result
An estimate from a reference point is useful when the function value at one input is known or easy to calculate, but the value at a nearby input is not. It gives a checkable guarantee without requiring the exact value at the second input. The estimate becomes tighter when the distance between inputs is smaller or when a smaller valid derivative bound is available.
Keep the two ingredients separate when working: first establish a bound on the derivative’s magnitude over the full interval; then multiply that bound by the distance between the inputs. Finally, add and subtract the resulting change bound from the known reference value. This order makes both the assumptions and the numerical estimate visible.
Check Your Understanding
For each question, identify the reference value, the interval, and the derivative bound needed before forming an estimate.
- Suppose \(f\) is continuous on \([1,1.2]\), differentiable in its interior, \(f(1)=4\), and \(|f'(t)|\le3\) throughout the interior. What interval is guaranteed to contain \(f(1.2)\)?
- For \(g(t)=t^2\), use \(g(2)=4\) and a valid bound for \(|g'|\) on \([1.9,2]\) to bound \(g(1.9)\).
- Why does the estimate use \(|x-a|\) rather than \(x-a\)?
- Does the condition \(f'(t)\le5\) alone guarantee that \(|f(x)-f(a)|\le5|x-a|\)? Explain.
- If a valid derivative bound is replaced by a larger constant, does the estimate remain guaranteed? How does the range typically change?
