Why the Base Point and Step Size Matter
A tangent-line estimate depends on two choices: the point where the tangent line is built and the distance from that point to the input being estimated. A base point with familiar function and derivative values can make the calculation easy. A base point closer to the target usually makes the estimate more reliable, because the function is being approximated over a shorter step.
These goals can compete. The easiest base point to calculate from may not be the nearest one. A good choice balances convenient information with a suitably short step, while respecting the function's domain and the estimates' units.
The sign and the distance have different roles. The signed step \(h\) determines whether the target lies to the right or left of \(a\), so it must be used in the linearization. The distance \(|h|\) is useful for assessing how far the estimate travels from its base point.
What Makes a Useful Base Point?
A practical base point should satisfy several conditions. First, \(f(a)\) and \(f'(a)\) should be known or straightforward to calculate. Second, \(a\) should be near the target \(x\), so that the step is short. Third, the function must be defined at the base point and differentiable there. If an error bound is being used, its assumptions must also hold on the interval between \(a\) and \(x\).
“Near” is not a fixed numerical threshold. A step of \(0.1\) may be small for one function and less reassuring for another. Curvature matters: a function that bends sharply can depart from its tangent line more rapidly than a function that bends gently. When a bound on the second derivative is available, it gives a precise way to account for that difference.
This result, established in the earlier material on linearization error bounds, says that the error is bounded by a constant times the square of the step distance. The number \(M\) must bound \(|f''|\) on the whole interval between the base point and target, not merely at one endpoint. The bound is an upper limit on the error, not necessarily the exact error.
How Changing the Step Changes the Bound
When the same second-derivative bound \(M\) applies over the relevant intervals, the bound changes quadratically with the step distance. The comparison follows directly from the theorem. If a new step has distance \(q\) times the original distance, where \(q\ge 0\), then its error bound is \(q^2\) times the original bound:
In particular, halving the step distance quarters the bound, provided the same \(M\) is valid for both intervals. This does not prove that the actual error is always reduced by exactly that factor. The theorem bounds the error; it does not specify its exact value. Also, if the intervals have different curvature bounds, compare the resulting bounds using their respective values of \(M\).
Prefer a point where the function value and derivative are accessible and where the function is differentiable.
Compute \(h=x-a\) for the tangent-line formula and \(|h|\) for judging how far the target is from the base.
A closer point may improve the error bound, but check whether its function and derivative values remain practical to use.
If \(|f''|\le M\) on the interval, calculate \(M|h|^2/2\) and describe it as a bound, not as the exact error.
Worked Examples
Worked Example: Comparing Two Base Points
Estimate \(f(2.1)\) for \(f(x)=x^2\), first using \(a=2\), then using \(a=1\). Compare the errors and their bounds.
For \(f(x)=x^2\), \(f'(x)=2x\) and \(f''(x)=2\). At \(a=2\), the base value is \(f(2)=2^2=4\), the slope is \(f'(2)=2(2)=4\), and the signed step is \(h=2.1-2=0.1\). Thus
The exact function value is \(f(2.1)=(2.1)^2=4.41\), so the actual absolute error is \(|4.41-4.4|=0.01\). Since \(|f''|=2\), take \(M=2\). The error bound is \((2/2)(0.1)^2=0.01\), which agrees with the actual error in this case.
Now use \(a=1\). Here \(f(1)=1\), \(f'(1)=2\), and \(h=2.1-1=1.1\). The estimate is
Its actual absolute error is \(|4.41-3.2|=1.21\). The bound is \((2/2)(1.1)^2=1.21\), again equal to the actual error. The base point \(a=2\) gives the better estimate because its step distance, \(0.1\), is much shorter than \(1.1\).
Check: Substituting each base point into \(f(a)+f'(a)(x-a)\) gives the estimates \(4.4\) and \(3.2\); direct evaluation confirms the target value is \(4.41\).
Worked Example: A Short Step and a Curvature Bound
Use \(f(x)=x^3\) and base point \(a=3\) to estimate \(f(3.02)\). Then use a second-derivative bound to check how close the estimate is allowed to be.
Differentiate: \(f'(x)=3x^2\), so \(f(3)=27\) and \(f'(3)=3(3^2)=27\). The signed step is \(h=3.02-3=0.02\). Therefore
For the bound, \(f''(x)=6x\). On \([3,3.02]\), this is increasing and positive, so \(|f''(x)|\le 6(3.02)=18.12\). We can take \(M=18.12\). Thus
Check against the exact value: \(f(3.02)=(3+0.02)^3=27+3(9)(0.02)+3(3)(0.02)^2+(0.02)^3=27+0.54+0.0036+0.000008=27.543608\). The actual error is \(27.543608-27.54=0.003608\), which is less than \(0.003624\), as the bound requires.
Check: The computed error is positive and is below the bound. The estimate is not the exact value, even though the step is small.
Worked Example: Balancing a Convenient Radius with a Closer One
The area of a circle of radius \(r\) is \(A(r)=\pi r^2\). Estimate the area at \(r=10.2\) cm using base points \(a=10\) cm and \(a=10.1\) cm, then compare the estimates.
The derivative is \(A'(r)=2\pi r\), with units of square centimetres per centimetre. At \(a=10\), \(A(10)=100\pi\) cm\(^2\), \(A'(10)=20\pi\) cm\(^2\)/cm, and \(h=10.2-10=0.2\) cm. The estimate is
At \(a=10.1\), \(A(10.1)=\pi(10.1)^2=102.01\pi\) cm\(^2\), \(A'(10.1)=20.2\pi\) cm\(^2\)/cm, and \(h=10.2-10.1=0.1\) cm. Hence
The exact area is \(A(10.2)=\pi(10.2)^2=104.04\pi\) cm\(^2\). The first estimate has absolute error \(0.04\pi\) cm\(^2\); the second has error \(0.01\pi\) cm\(^2\). Because \(A''(r)=2\pi\), the error bound is \(\pi h^2\). It gives \(0.04\pi\) cm\(^2\) at \(a=10\) and \(0.01\pi\) cm\(^2\) at \(a=10.1\), matching the errors here. The closer base point halves the distance and quarters the bound.
Check: The direct calculation \(A(10.2)=104.04\pi\) confirms both reported errors by subtraction. Each estimate has units of area.
A Common Pitfall: Using Distance as the Signed Step
The distance \(|x-a|\) is always nonnegative, but the tangent-line formula needs the signed step \(x-a\). Replacing the signed step with the distance can reverse the direction of the estimate when the target lies to the left of the base point.
Worked Example: Correcting a Sign Error
For \(f(x)=x^2\), estimate \(f(1.9)\) from \(a=2\). Since \(f(2)=4\), \(f'(2)=4\), and \(h=1.9-2=-0.1\), the correct estimate is
A wrong calculation uses the distance \(0.1\) in place of the signed step: \(4+4(0.1)=4.4\). This treats the target as though it were to the right of \(2\). In fact, \(f(1.9)=(1.9)^2=3.61\), so the correct estimate has error \(|3.61-3.6|=0.01\); the wrong estimate has error \(|3.61-4.4|=0.79\).
Check: Substitution into \(L_a(x)=f(a)+f'(a)(x-a)\) requires \(1.9-2=-0.1\), confirming the corrected estimate.
Choosing with Care
A nearby base point is a useful starting preference, not an automatic guarantee that every estimate will be more accurate. The actual error also depends on the function's behavior between the base and target. A second-derivative bound gives a controlled comparison when its conditions hold; without such a bound, closeness alone is a reason to expect a more local estimate, not a quantified promise.
When comparing candidate base points, keep the calculation organized: record each point's known function value, slope, signed step, and step distance. A simple table can prevent a convenient value from hiding an unnecessarily long step.
| Quantity | Question to ask |
|---|---|
| Base value \(f(a)\) | Is it known or easy to calculate? |
| Slope \(f'(a)\) | Can it be found reliably at this point? |
| Signed step \(x-a\) | Does it point from the base point toward the target? |
| Distance \(|x-a|\) | Is this candidate substantially closer than the alternatives? |
| Curvature bound \(M\) | Is \(|f''|\le M\) valid across the entire interval used? |
Check Your Understanding
Answer these questions using the base-point and step-size ideas.
- For a target \(x=4.08\), compare the step distances from base points \(a=4\) and \(a=3.5\). Which is closer?
- If the same value of \(M\) applies and the step distance is reduced to one third of its original value, what happens to the second-derivative error bound?
- Why must \(x-a\), rather than \(|x-a|\), appear in the linearization formula?
- What interval must a proposed value of \(M\) cover for the error bound to apply?
- Does an error bound equal the actual error in every linear estimate? Explain briefly.
