Choosing a Base Point
A linear approximation uses the tangent line to a function at a chosen base point to estimate a nearby function value. The approximation is most useful when two tasks are manageable: evaluating the function and its derivative at the base point, and computing the change from that point to the target.
These tasks can pull in different directions. A point very close to the target may require awkward arithmetic, while a slightly more distant point may have a function value and derivative that are much easier to calculate. Choosing effectively means comparing the full calculation, not simply picking the nearest familiar number.
The formula requires that \(f(a)\) and \(f'(a)\) exist. For a useful estimate, \(a\) should also be reasonably near \(b\), and the function should behave sufficiently regularly between them. The formula itself does not guarantee a particular numerical accuracy: choosing a good point makes a sensible first-order estimate, but the size of the actual error depends on the function.
What Makes a Point Effective?
Begin by identifying the target input \(b\). Then consider candidate base points \(a\) using several questions:
- Is it near the target? The change \(b-a\) is the step taken along the tangent line. A smaller step is generally desirable, all else being equal.
- Can you compute \(f(a)\) exactly or simply? The value at the base point is part of the estimate, so an awkward value can make the whole calculation less useful.
- Can you compute \(f'(a)\) simply? A simple derivative value reduces arithmetic and the chance of error.
- Is the function defined and differentiable there? A point outside the domain, or at a point where the derivative does not exist, cannot serve as a base point for this linearization.
These are practical criteria, not a theorem that names one universally best point. Two candidates can each have an advantage. Compare the resulting linear estimates and use the context of the problem to decide whether the simpler calculation justifies a somewhat larger step.
Why the Linearization Is a Local Estimate
The reason for preferring a nearby point comes from the definition of the derivative. If \(f\) is differentiable at \(a\), then for a small change \(h\), the difference between \(f(a+h)-f(a)\) and \(f'(a)h\) is small relative to \(h\). More precisely, the remainder \(r(h)\) defined by \(f(a+h)=f(a)+f'(a)h+r(h)\) satisfies \(r(h)/h\to 0\) as \(h\to 0\). Thus the linearization keeps the function value and derivative terms and omits a remainder that becomes small relative to the step.
To estimate \(f(b)\), take \(h=b-a\). The linearization is the part \(f(a)+f'(a)h\), so it gives \(L_a(b)\). This explains why proximity matters, but it does not say that the closest available point always produces the most accurate estimate. The omitted remainder depends on the function, and the ease of carrying out the calculation matters too.
Worked Comparisons
Worked Example: Choosing between a nearby point and a distant point
Estimate \(f(2.1)\) for \(f(x)=x^2\), comparing base points \(a=2\) and \(a=1\). Since \(f'(x)=2x\), the linearization at a base point \(a\) is \(L_a(x)=a^2+2a(x-a)\).
For \(a=2\), the base value is \(f(2)=2^2=4\), the slope is \(f'(2)=2(2)=4\), and the step to the target is \(2.1-2=0.1\). Therefore, $$L_2(2.1)=4+4(0.1)=4+0.4=4.4.$$
For \(a=1\), the base value is \(f(1)=1^2=1\), the slope is \(f'(1)=2(1)=2\), and the step is \(2.1-1=1.1\). Therefore, $$L_1(2.1)=1+2(1.1)=1+2.2=3.2.$$
For a check, the target value is \(f(2.1)=(2.1)^2=4.41\). The estimate from \(a=2\) is \(4.4\), whereas the estimate from \(a=1\) is \(3.2\). The first is much closer to the target value, and it also uses the smaller step. Here \(a=2\) is the effective choice.
Worked Example: Compare two convenient base points
Estimate \(f(2.3)\) for \(f(x)=(x-2)^2+5\), comparing \(a=2\) and \(a=2.5\). The derivative is \(f'(x)=2(x-2)\).
At \(a=2\), \(f(2)=(2-2)^2+5=5\) and \(f'(2)=2(2-2)=0\). The step is \(2.3-2=0.3\), so $$L_2(2.3)=5+0(0.3)=5.$$
At \(a=2.5\), \(f(2.5)=(2.5-2)^2+5=(0.5)^2+5=5.25\), and \(f'(2.5)=2(2.5-2)=1\). The step is \(2.3-2.5=-0.2\), so $$L_{2.5}(2.3)=5.25+1(-0.2)=5.05.$$
Check against the direct value: $$f(2.3)=(2.3-2)^2+5=(0.3)^2+5=5.09.$$
The estimates are \(5\) and \(5.05\); the second is closer to \(5.09\). Both points are straightforward to use, and \(2.5\) is nearer the target than \(2\). In this comparison, the nearer base point gives the closer estimate.
Worked Example: A nearby point can still give a poor estimate
Consider \(f(x)=x^3-3x\) and estimate \(f(1.1)\) using \(a=1\) and \(a=1.5\). The derivative is \(f'(x)=3x^2-3\).
At \(a=1\), \(f(1)=1^3-3(1)=-2\), and \(f'(1)=3(1)^2-3=0\). Since \(1.1-1=0.1\), $$L_1(1.1)=-2+0(0.1)=-2.$$
At \(a=1.5\), \(f(1.5)=(1.5)^3-3(1.5)=3.375-4.5=-1.125\), and $$f'(1.5)=3(1.5)^2-3=3(2.25)-3=6.75-3=3.75.$$
The step from \(1.5\) to \(1.1\) is \(1.1-1.5=-0.4\), so $$L_{1.5}(1.1)=-1.125+3.75(-0.4)=-1.125-1.5=-2.625.$$
Check against the direct target value: $$f(1.1)=(1.1)^3-3(1.1)=1.331-3.3=-1.969.$$
The estimate \(-2\), from the closer point \(a=1\), is much nearer \(-1.969\) than \(-2.625\). In this case the derivative at \(1\) is zero, making its linearization particularly simple, while the other candidate produces a steep tangent line and a larger change over its step. Distance alone does not settle the comparison.
A Practical Selection Process
When a problem offers several plausible points—or does not specify one—use a deliberate comparison rather than choosing by appearance.
Write down the input \(b\) whose function value is required.
Look for points near \(b\) where the function is defined and differentiable.
For each candidate \(a\), consider how simply you can find \(f(a)\), \(f'(a)\), and \(b-a\).
Compute \(L_a(b)=f(a)+f'(a)(b-a)\) for the strongest candidates, then choose the estimate best supported by closeness and computational simplicity.
A convenient point is not automatically a suitable point. It must lie where the function and derivative are defined. For instance, if a formula is only defined on part of the real line, a numerically close candidate outside that domain is not available. Likewise, if the derivative does not exist at a candidate, the tangent-line linearization there is not defined.
Common Mistake: Using the Target Instead of the Step
The linearization uses the change from the base point, \(b-a\), not the target input \(b\) by itself. In the first worked example, using \(f(2)=4\) and \(f'(2)=4\) but multiplying the slope by \(2.1\) gives the incorrect calculation $$4+4(2.1)=12.4.$$
This calculation treats the change as \(2.1\), even though the base point is \(2\). The correct change is \(2.1-2=0.1\), so the correct estimate is $$4+4(2.1-2)=4+4(0.1)=4.4.$$
Check Your Understanding
Use the linearization formula and the selection criteria to answer these questions.
- For a target input \(b\), what three quantities are needed to evaluate the linearization at a candidate base point \(a\)?
- Why must a candidate base point be a point where the function is differentiable?
- For \(f(x)=x^2\) and target \(b=3.1\), compare the steps \(b-a\) for \(a=3\) and \(a=2\). Which candidate is closer?
- Does choosing the closest candidate guarantee the most accurate estimate? Explain using the role of the omitted remainder.
- In the first worked example, why is \(4+4(2.1)\) not the linearization estimate at \(a=2\)?
