From a Finite Change to an Intermediate Derivative
A finite difference quotient measures the change in a function over a nonzero interval. It is often used as a numerical estimate of a derivative, which describes change at a single point. The Mean Value Theorem connects these two quantities exactly: under suitable hypotheses, the quotient equals the derivative at some point between the endpoints.
This exact relationship is useful even when the intermediate point is not known. If we know that the derivative stays within certain bounds, then the quotient must satisfy those same bounds. If we also know a bound on the second derivative, we can quantify how far the quotient may be from the derivative at the starting point.
The Exact Mean Value Theorem Relationship
The Mean Value Theorem says that if \(f\) is continuous on a closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there is a point \(c\in(a,b)\) such that \(f'(c)=(f(b)-f(a))/(b-a)\). Applying it to the interval with endpoints \(x\) and \(x+h\) gives the finite-difference version below. The theorem applies whether \(h\) is positive or negative.
The point \(c\) is guaranteed to exist, but the theorem does not usually identify it explicitly. The quotient is therefore not necessarily equal to \(f'(x)\), nor to \(f'(x+h)\). Rather, it matches the derivative at at least one intermediate point.
Worked Example: Finding the Intermediate Point
Let \(f(t)=t^2\), with \(x=2\) and \(h=3\). The endpoints are \(2\) and \(5\), and \(f\) is continuous and differentiable on the required intervals.
First compute the quotient: $$ Q=\frac{f(2+3)-f(2)}{3} =\frac{f(5)-f(2)}{3} =\frac{25-4}{3} =\frac{21}{3} =7. $$ Since \(f'(t)=2t\), the intermediate point must satisfy \(2c=7\), so \(c=7/2\). This point is strictly between \(2\) and \(5\).
Check the result by substituting \(c=7/2\) into the derivative: $$ f'(7/2)=2(7/2)=7=Q. $$ Thus the quotient exactly equals the derivative at the intermediate point.
Using Derivative Bounds to Bound the Quotient
Suppose the derivative has a lower bound \(m\) and an upper bound \(M\) throughout the interval between the endpoints. The Mean Value Theorem gives \(Q=f'(c)\) for some point \(c\) in that interval. Therefore the same bounds apply to \(Q\). This conclusion does not require the derivative to be monotone; it only requires the bounds to hold at every interior point.
Worked Example: Bounding a Quotient by Derivative Values
Let \(f(t)=t^3\), \(x=1\), and \(h=1\). Then \(f'(t)=3t^2\). For \(1<t<2\), we have \(3<3t^2<12\), so the derivative is bounded below by \(3\) and above by \(12\) on the interval. The derivative-bound rule therefore gives \(3\le Q\le12\).
Compute the quotient to check: $$ Q=\frac{f(2)-f(1)}{1} =\frac{8-1}{1} =7. $$ Indeed, \(3\le7\le12\). The exact intermediate point also exists: \(3c^2=7\), so \(c=\sqrt{7/3}\), which lies between \(1\) and \(2\). Substitution verifies \(f'(c)=3(7/3)=7=Q\).
Bounding the Error as an Estimate of the Starting Derivative
A common use of the quotient is to estimate \(f'(x)\). To control this error, suppose the second derivative is bounded in absolute value. It is important to state endpoint regularity carefully: the argument below applies when \(f\) is continuously differentiable on the closed interval and twice differentiable in its interior. These hypotheses ensure that the needed endpoint derivative exists and that Rolle’s Theorem can be applied in the proof.
The factor \(1/2\) comes from a second-order remainder for the function values. Dividing that remainder by the step size converts a bound proportional to \(h^2\) into a bound proportional to \(|h|\) for the quotient.
Write \(R=f(x+h)-f(x)-f'(x)h\). This is the difference between the actual function change and the change predicted using the derivative at \(x\).
Consider \(g(t)=f(t)-f(x)-f'(x)(t-x)-\frac{R}{h^2}(t-x)^2\) on the interval between \(x\) and \(x+h\). It has \(g(x)=g(x+h)=0\) and \(g'(x)=0\). Rolle’s Theorem gives a zero of \(g'\) strictly between the endpoints; applying Rolle’s Theorem to \(g'\) between that zero and \(x\) gives an interior point \(\eta\) with \(g''(\eta)=0\). The stated regularity ensures these applications are valid, including when \(x\) is an endpoint.
Since \(g''(t)=f''(t)-2R/h^2\), the point \(\eta\) satisfies \(R=\frac12 f''(\eta)h^2\). Hence \(|R|\le K h^2/2\). Finally, \(Q-f'(x)=R/h\), so \(|Q-f'(x)|\le K|h|/2\).
This proof also explains the role of each hypothesis. The continuity and differentiability of \(f\) make the Rolle applications legitimate; the second derivative need not be continuous, but it must exist at the interior points used. A bound on \(f''\) then controls the remainder.
Worked Example: A Sharp Quotient-Error Bound
Let \(f(t)=t^2\), \(x=1\), and \(h=1/2\). On the interval \([1,3/2]\), \(f'(t)=2t\) is continuous, \(f''(t)=2\), and we may take \(K=2\). The theorem predicts $$ |Q-f'(1)|\le \frac{2(1/2)}{2}=\frac12. $$ Compute the quotient: $$ Q=\frac{f(3/2)-f(1)}{1/2} =\frac{9/4-1}{1/2} =\frac{5/4}{1/2} =\frac52. $$ Since \(f'(1)=2\), the actual error is $$ |Q-f'(1)|=\left|\frac52-2\right|=\frac12. $$ It equals the bound. The check is consistent with the exact intermediate derivative: \(f'(5/4)=2(5/4)=5/2=Q\).
Worked Example: A Nonlinear Function with a Strict Error Bound
Take \(f(t)=1/t\), \(x=1\), and \(h=1/10\). On \([1,11/10]\), the function is continuously differentiable and twice differentiable in the interior. Its derivatives are \(f'(t)=-1/t^2\) and \(f''(t)=2/t^3\). Thus \(|f''(t)|\le2\), so \(K=2\).
The quotient is $$ Q=\frac{f(11/10)-f(1)}{1/10} =\frac{10/11-1}{1/10} =\frac{-1/11}{1/10} =-\frac{10}{11}. $$ The derivative at the starting point is \(f'(1)=-1\), and the error is $$ |Q-f'(1)| =\left|-\frac{10}{11}+1\right| =\frac{1}{11}. $$ The theorem gives the bound \(K|h|/2=2(1/10)/2=1/10\). Since \(1/11<1/10\), the computed error satisfies the bound. As an additional check of the exact Mean Value Theorem relationship, \(f'(c)=-10/11\) implies \(c^2=11/10\), so \(c=\sqrt{11/10}\), which lies strictly between \(1\) and \(11/10\).
A Common Error: Forgetting to Divide by the Step
The second-order remainder controls a difference in function values: $$ |f(x+h)-f(x)-f'(x)h|\le \frac{K h^2}{2}. $$ But a finite difference quotient divides the function change by \(h\). Therefore the corresponding quotient error is bounded by \(K|h|/2\), not by \(Kh^2/2\).
Worked Example: Correcting a Missing Division
For \(f(t)=t^2\), \(x=1\), \(h=1/2\), and \(K=2\), a mistaken calculation might claim that the quotient error is at most \(Kh^2/2=2(1/4)/2=1/4\). But the actual quotient is \(5/2\), while \(f'(1)=2\), so the error is \(1/2\). The proposed bound \(1/4\) is false.
The function-value remainder bound is \(Kh^2/2=1/4\). To obtain the quotient error, divide by \(|h|=1/2\): $$ \frac{1/4}{1/2}=\frac12=\frac{K|h|}{2}. $$ This agrees with the actual error. Always distinguish a bound on the function-value remainder from a bound on the difference quotient’s error.
Choosing the Step Size
The error estimate shows how the step size affects the accuracy of the quotient as an estimate of \(f'(x)\). If \(K\) is fixed, reducing \(|h|\) reduces the guaranteed error in direct proportion. For example, replacing a step of size \(0.1\) by one of size \(0.01\) reduces this theoretical bound by a factor of ten.
The result is a guarantee under the stated assumptions, not a promise that every smaller step produces a smaller error in every numerical computation. In numerical work, rounding can also affect computed values; the theorem concerns the exact quotient. The central distinction remains: the Mean Value Theorem gives an exact derivative value somewhere inside the interval, while a second-derivative bound controls how far that quotient can be from the derivative at the chosen starting point.
Check Your Understanding
Use the Mean Value Theorem and the bounds stated in each question where appropriate.
- For \(f(t)=t^2\), \(x=3\), and \(h=2\), compute the quotient and find an intermediate point \(c\) for which \(Q=f'(c)\).
- If \(4\le f'(t)\le9\) throughout the interval between \(x\) and \(x+h\), what bounds must the quotient satisfy?
- Suppose \(|f''(t)|\le6\) on the relevant interval and \(|h|=0.2\). What bound follows for \(|Q-f'(x)|\)?
- What regularity assumptions in the error-bound theorem ensure the endpoint derivative and the Rolle’s Theorem argument are valid?
- If the function-value remainder is bounded by \(Kh^2/2\), what is the corresponding bound for the quotient error, and why?
