From a Nearby Input Change to a Function-Value Error
A small change in input often produces a change in function value that can be estimated using the derivative at a nearby base point. If the input changes from \(a\) to \(a+h\), the derivative-based estimate of the function’s change is \(f'(a)h\). This tutorial focuses on what that estimate means and how to calculate its error.
The key is to compare like with like. The quantity \(f'(a)h\) estimates the change \(f(a+h)-f(a)\), not the new function value \(f(a+h)\) itself. To estimate the new value, add the estimated change to \(f(a)\).
The sign of the signed error tells whether the actual change is above or below the estimate. The absolute error measures the size of the discrepancy without regard to its direction.
Why the Derivative Gives a First-Order Estimate
At a point \(a\), differentiability means that the difference quotient for \(f\) approaches a finite limit as the input change approaches zero. In the notation of the derivative, this says that for nonzero \(h\) sufficiently close to zero, the quotient \(\frac{f(a+h)-f(a)}{h}\) approaches \(f'(a)\).
Rearranging the difference quotient shows why multiplying the derivative by \(h\) estimates the function change. The complete argument also describes how the approximation error behaves.
By the definition of \(f'(a)\), the quantity \(\frac{f(a+h)-f(a)}{h}-f'(a)\) approaches zero as \(h\) approaches zero through nonzero values.
For nonzero \(h\), the signed error equals \(h\left(\frac{f(a+h)-f(a)}{h}-f'(a)\right)\). This is exactly the actual change minus \(f'(a)h\).
The absolute error is \(|h|\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|\). The second factor approaches zero, so the absolute error divided by \(|h|\) approaches zero.
Thus, differentiability does more than provide a formula: it guarantees that the error becomes small compared with the size of the input change as \(h\) approaches zero. This is an asymptotic statement. By itself, it does not give a particular numerical error bound for a chosen nonzero \(h\); to calculate the error at a specific input, compare the actual change with the estimate.
Worked Example: A Quadratic Near Its Base Point
Worked Example: A Quadratic Near Its Base Point
Let \(f(x)=x^2\), with base point \(a=3\) and input change \(h=0.1\). Find the actual change, the first-order estimate, and the absolute approximation error.
First, the new input is \(a+h=3+0.1=3.1\). The function values are \(f(3)=3^2=9\) and \(f(3.1)=(3.1)^2=9.61\). Therefore the actual change is \(9.61-9=0.61\).
The derivative is \(f'(x)=2x\), so \(f'(3)=6\). The first-order estimate of the change is \(f'(3)h=6(0.1)=0.6\).
The signed error, actual change minus estimate, is \(0.61-0.6=0.01\). Its absolute value is \(0.01\). As a check, the new function value estimated by the linearization is \(f(3)+0.6=9.6\), which differs from the actual value \(9.61\) by \(0.01\).
Worked Example: A Reciprocal Function
Worked Example: A Reciprocal Function
Let \(f(x)=\frac{1}{x}\), with \(a=2\) and \(h=0.1\). Calculate the actual change, the first-order estimate, and the absolute error.
The new input is \(2+0.1=2.1=\frac{21}{10}\). Hence \(f(2)=\frac12\) and \(f(2.1)=\frac{10}{21}\). The actual change is \(\frac{10}{21}-\frac12=\frac{20}{42}-\frac{21}{42}=-\frac{1}{42}\).
Since \(f'(x)=-\frac{1}{x^2}\), we have \(f'(2)=-\frac14\). The first-order estimate is \(-\frac14(0.1)=-\frac14\cdot\frac{1}{10}=-\frac{1}{40}\).
The signed error is \(-\frac{1}{42}-(-\frac{1}{40})=-\frac{20}{840}+\frac{21}{840}=\frac{1}{840}\). Therefore the absolute error is also \(\frac{1}{840}\), approximately \(0.00119048\) to eight decimal places.
To check, the estimated new function value is \(\frac12-\frac{1}{40}=\frac{19}{40}\). The difference between the actual value and this estimate is \(\frac{10}{21}-\frac{19}{40}=\frac{400}{840}-\frac{399}{840}=\frac{1}{840}\), as required.
Worked Example: A Negative Input Change
Worked Example: A Negative Input Change
Let \(f(x)=x^3\), \(a=1\), and \(h=-0.1\). Find the actual change, its first-order estimate, and the absolute error.
The new input is \(1-0.1=0.9\). Since \(f(1)=1\) and \(f(0.9)=(0.9)^3=0.729\), the actual change is \(0.729-1=-0.271\).
The derivative is \(f'(x)=3x^2\), so \(f'(1)=3\). The estimated change is \(3(-0.1)=-0.3\). The signed error is \(-0.271-(-0.3)=0.029\), and the absolute error is \(0.029\).
As a check, adding the estimated change to the base value gives the estimated function value \(1-0.3=0.7\). The actual value \(0.729\) differs from \(0.7\) by \(0.029\). The negative input change makes the estimated change negative, but it does not make the absolute error negative.
Calculating and Interpreting the Error
For a particular function and input change, a reliable calculation follows the same sequence: find the actual change, find the derivative-based estimate, then subtract the estimate from the actual change. Take the absolute value only after forming that difference.
If the goal is instead to estimate the function value at the new input, the estimate is \(f(a)+f'(a)h\). Its absolute error is the same quantity, because \(f(a+h)-\bigl(f(a)+f'(a)h\bigr)=f(a+h)-f(a)-f'(a)h\).
Units can help keep the interpretation clear. If \(x\) is measured in metres and \(f(x)\) in seconds, then \(f'(a)\) has units of seconds per metre. Multiplying by \(h\), measured in metres, gives \(f'(a)h\) in seconds, the same units as a change in \(f\). The absolute approximation error also has units of seconds.
A Common Mistake: Comparing the Wrong Quantities
For the reciprocal example, the actual change was \(-\frac{1}{42}\), while its estimate was \(-\frac{1}{40}\). A common mistake is to report \(\left|-\frac{1}{40}\right|=\frac{1}{40}\) as the approximation error. That is the size of the estimate, not the difference between the estimate and the actual change.
The correct calculation compares the two changes: \(\left|-\frac{1}{42}-(-\frac{1}{40})\right|=\frac{1}{840}\). Alternatively, compare function values: the actual value is \(\frac{10}{21}\), the estimated value is \(\frac{19}{40}\), and their absolute difference is \(\left|\frac{10}{21}-\frac{19}{40}\right|=\frac{1}{840}\). Both checks give the same error.
When the Estimate Applies
The first-order estimate uses the derivative at the chosen base point \(a\). For its justification, \(f\) must be differentiable at \(a\); to compute the actual error for a specific \(h\), the function must also be defined at \(a+h\). A derivative value alone does not state the exact change for a finite \(h\). The estimate is most useful when \(h\) is close to zero, and the exact error can be assessed by direct calculation when the function value is available.
Keep the roles of the quantities distinct: \(f'(a)h\) estimates a change, \(f(a)+f'(a)h\) estimates the new function value, and the absolute approximation error is the magnitude of the difference between the actual and estimated changes (or, equivalently, the actual and estimated new values).
Check Your Understanding
For each question, keep the actual change, estimated change, and absolute error distinct.
- For \(f(x)=x^2\), \(a=2\), and \(h=0.05\), what are the actual change and the estimate \(f'(a)h\)? What is the absolute error?
- For \(f(x)=\frac{1}{x}\), \(a=1\), and \(h=-0.1\), write the expression for the absolute approximation error before evaluating it.
- Why does \(f'(a)h\) estimate a change rather than the new function value?
- State what differentiability at \(a\) implies about absolute approximation error relative to \(|h|\) as \(h\) approaches zero.
- If an estimated change is negative, must its absolute approximation error be negative? Explain using the definition.
