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Linearization Workshop with Radicals and Rational Functions

Learn to choose a useful base point, compute a linear estimate for common radicals and rational functions, and assess its accuracy against the exact value.

Intermediate 9 min read

What You'll Learn

  • Write the linearization of a differentiable function at a chosen base point.
  • Select nearby base points where both the function value and derivative are manageable.
  • Linearize square-root and cube-root functions while respecting their domains.
  • Linearize a rational function and compare its estimate with an exact fraction.
  • Check the size and direction of approximation errors using exact-value calculations.

Introduction

A complicated function can sometimes be estimated near an input where its value is easy to calculate. The main idea is to replace the function, close to that input, with its tangent line. This workshop applies that idea to square roots, cube roots, and rational functions, then checks the resulting estimates against exact values.

A good estimate depends on two choices: the function’s base point and how close the requested input is to that point. A convenient base point makes the calculation simpler, but the final comparison with the exact value tells us how well the approximation performs in a particular case.

The Linearization Formula

Definition: If \(f\) is differentiable at \(a\), its linearization at \(a\) is the affine function \(L(x)=f(a)+f'(a)(x-a)\). The value \(L(x)\) is used to estimate \(f(x)\) when \(x\) is close to \(a\).

The formula uses the function’s value and slope at the base point. In a graph, \(L\) is the tangent line to \(y=f(x)\) at \(x=a\). The input change is \(x-a\); the estimated output change is \(f'(a)(x-a)\).

Before using the formula, check that \(f(a)\) and \(f'(a)\) exist. For radicals, also check the real-valued domain. For rational functions, the denominator must be nonzero at the base point.

Why the Tangent Line Gives a Local Estimate

Differentiability at \(a\) means that the change in \(f\) can be written as its linear change plus a remainder that becomes small compared with the input change. Specifically, there is a remainder \(r(h)\) such that

$$ f(a+h)=f(a)+f'(a)h+r(h), \qquad \lim_{h\to 0}\frac{r(h)}{h}=0. $$

Here \(h=x-a\). Since the first two terms on the right are \(L(a+h)\), the difference between the function and its linearization is \(r(h)\). The limit says that this difference, divided by the input change, tends to zero as the input approaches the base point. This is why the tangent line provides a first-order local estimate. It does not say that the estimate is exact away from the base point, or that the error has a particular sign.

Takeaway: Linearization gives a local estimate, not an identity. The closer \(x\) is to \(a\), the more relevant the differentiability limit is; checking an actual estimate still requires comparing it with \(f(x)\).

Choosing a Base Point

A useful base point is close to the input of interest, lies in the function’s domain, and makes both \(f(a)\) and \(f'(a)\) easy to compute. Familiar perfect powers are convenient for radicals: \(4\) for square roots and \(8\) for cube roots. For a rational function such as \(1/x\), a nearby nonzero integer can make the function value and derivative straightforward.

The choice is a balance, not a guarantee of a particular accuracy. A simple base point that is very far from the requested input may be less useful than a nearby one with slightly more arithmetic. In each example, we will calculate the linearization, use it at the target input, and check the result against the exact value.

Worked Examples

Worked Example: Estimating a Square Root

Estimate \(\sqrt{4.1}\) by linearizing \(f(x)=\sqrt{x}\) at \(a=4\), and compare the estimate with the exact value.

The function is real-valued for \(x\geq 0\), and its derivative at positive \(x\) is \(f'(x)=1/(2\sqrt{x})\). Thus

$$ f(4)=2, \qquad f'(4)=\frac{1}{2\sqrt{4}}=\frac14. $$

The linearization is

$$ L(x)=2+\frac14(x-4). $$

At \(x=4.1\), the input change is \(4.1-4=0.1\), so

$$ L(4.1) =2+\frac14(0.1) =2+0.025 =2.025. $$

For an exact check, \(2.025=81/40\), and

$$ (2.025)^2 =\left(\frac{81}{40}\right)^2 =\frac{6561}{1600} =4.100625. $$

Since \(4.100625>4.1\), the estimate \(2.025\) is slightly greater than \(\sqrt{4.1}\). More precisely, the difference is

$$ 2.025-\sqrt{4.1} = \frac{(2.025)^2-4.1}{2.025+\sqrt{4.1}} = \frac{0.000625}{2.025+\sqrt{4.1}} \approx 0.000154. $$

Thus \(\sqrt{4.1}\approx 2.024846\) to six decimal places, while the linear estimate is \(2.025\). The computed sign and size of the difference agree with the squared-value check.

Worked Example: Estimating a Cube Root

Estimate \(\sqrt[3]{8.3}\) using the base point \(a=8\), then check the estimate against the exact value.

For \(f(x)=\sqrt[3]{x}\), the derivative at nonzero \(x\) is \(f'(x)=1/(3x^{2/3})\). At \(8\),

$$ f(8)=2, \qquad f'(8)=\frac{1}{3\cdot 8^{2/3}} =\frac{1}{3\cdot 4} =\frac{1}{12}. $$

Therefore,

$$ L(x)=2+\frac{1}{12}(x-8). $$

The input change from \(8\) to \(8.3\) is \(0.3\), giving

$$ L(8.3) =2+\frac{0.3}{12} =2+0.025 =2.025. $$

To check, cube the estimate. Since \(2.025=81/40\),

$$ (2.025)^3 =\left(\frac{81}{40}\right)^3 =\frac{531441}{64000} =8.303765625. $$

This is greater than \(8.3\), so the estimate is above the cube root. The difference can also be expressed exactly in terms of the root:

$$ 2.025-\sqrt[3]{8.3} = \frac{(2.025)^3-8.3} {(2.025)^2+2.025\sqrt[3]{8.3}+(\sqrt[3]{8.3})^2}. $$

The numerator is \(8.303765625-8.3=0.003765625\). Evaluating the root gives \(\sqrt[3]{8.3}\approx2.024694\) to six decimal places, so the difference is about \(0.000306\). Cubing \(2.024694\) gives approximately \(8.300002\), consistent with that rounded root and with the exact-value comparison.

Worked Example: Estimating a Rational Function

Estimate \(1/4.1\) by linearizing \(f(x)=1/x\) at \(a=4\), and compare the estimate with the exact fraction.

The domain excludes \(x=0\). On that domain,

$$ f'(x)=-\frac{1}{x^2}, \qquad f(4)=\frac14, \qquad f'(4)=-\frac{1}{16}. $$

The linearization is

$$ L(x)=\frac14-\frac{1}{16}(x-4). $$

At \(x=4.1\),

$$ L(4.1) =\frac14-\frac{0.1}{16} =0.25-0.00625 =0.24375 =\frac{39}{160}. $$

The exact value is \(1/4.1=10/41\). Subtracting the exact value from the estimate gives

$$ \frac{39}{160}-\frac{10}{41} = \frac{39\cdot41-10\cdot160}{160\cdot41} = \frac{1599-1600}{6560} = -\frac{1}{6560} \approx -0.000152. $$

The negative difference shows that the linear estimate is below the exact value. In decimals, \(10/41\approx0.243902\) to six decimal places, compared with the estimate \(0.243750\). The exact fraction confirms both the direction and approximate size of the error.

A Common Mistake: Losing the Sign of the Slope

For \(f(x)=1/x\) at \(a=4\), the slope is \(-1/16\), not \(1/16\). A sign error leads to the wrong linearization:

$$ \text{Incorrect:}\qquad L_{\mathrm{wrong}}(x)=\frac14+\frac{1}{16}(x-4). $$

At \(x=4.1\), this gives

$$ L_{\mathrm{wrong}}(4.1) =\frac14+\frac{0.1}{16} =0.25625. $$

But the exact value \(1/4.1\approx0.243902\) is less than \(1/4=0.25\), because increasing a positive input decreases its reciprocal. The incorrect estimate moves in the opposite direction. Using \(f'(4)=-1/16\) gives \(0.24375\), consistent with the function’s decreasing behavior and close to the exact value.

A useful check is therefore to ask whether the estimate changes in the direction indicated by the derivative. This check can catch a sign error, though it does not replace calculating the linearization and comparing with the exact value.

Comparing the Three Estimates

Function and target Base point Linear estimate Exact-value check
\(\sqrt{4.1}\) \(4\) \(2.025\) About \(2.024846\); estimate is high
\(\sqrt[3]{8.3}\) \(8\) \(2.025\) About \(2.024694\); estimate is high
\(1/4.1\) \(4\) \(0.243750\) About \(0.243902\); estimate is low

The table illustrates why an estimate should be checked rather than assumed exact. In the radical examples the tangent-line estimate lies above the exact value; in the reciprocal example it lies below. The linearization formula alone does not determine the error’s direction.

Why These Calculations Matter

Linear estimates are useful when a function value is inconvenient to compute directly but the value and derivative at a nearby base point are simple. The method also makes the role of the slope explicit: a small input change is converted into an estimated output change by multiplying by the derivative at the base point.

For reliable use, keep the checks together: the base point must be in the domain, the derivative must exist there, the displacement must be calculated with its sign, and the resulting estimate should be compared with the exact value when that value is available. A well-chosen base point simplifies the work; the exact-value comparison shows what the approximation achieved.

Check Your Understanding

For each question, write down the base point, derivative, linear estimate, and a check against the exact value where requested.

  1. Linearize \(f(x)=\sqrt{x}\) at \(a=9\), then estimate \(\sqrt{9.2}\).
  2. For \(f(x)=\sqrt[3]{x}\), find the linearization at \(a=27\), then estimate \(\sqrt[3]{26.7}\).
  3. Linearize \(f(x)=1/x\) at \(a=2\), then estimate \(1/2.1\). Is the estimate above or below the exact value?
  4. Why must the base point for \(f(x)=1/x\) be nonzero?
  5. For a differentiable function at \(a\), what does the limit of the remainder divided by the input change say about linearization?

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