Linearization as a Tangent-Line Approximation

From a Curve to a Tangent-Line Estimate

A differentiable function can be complicated to evaluate directly, while its tangent line at a nearby point is simple. Near the point of tangency, that line often gives a useful estimate of the function's value. The estimate is called a linearization.

The key is to use the function's value and slope at the chosen center. The resulting line agrees with the function at that point and has the same slope there. Its usefulness depends on evaluating close to the center: a tangent line is a local model, not a replacement for the function everywhere.

Definition of Linearization

Let \(f\) be differentiable at \(a\). The degree-one Taylor polynomial of \(f\) centered at \(a\) is

$$ P_1(x)=f(a)+f'(a)(x-a). $$

This polynomial, viewed as a line, is the linearization of \(f\) at \(a\). The value \(a\) is its center. To estimate \(f(x)\) for \(x\) near \(a\), use \(P_1(x)\) in place of \(f(x)\).

Definition: If \(f\) is differentiable at \(a\), its linearization at \(a\) is \(L(x)=f(a)+f'(a)(x-a)\). It is the equation of the tangent line to \(y=f(x)\) at the point \((a,f(a))\).

The form \(f(a)+f'(a)(x-a)\) records two facts: the line passes through \((a,f(a))\), and its slope is \(f'(a)\). In particular, substituting \(x=a\) gives \(L(a)=f(a)\). The degree-one Taylor polynomial is thus not an arbitrary convenient line; it is precisely the tangent-line model.

Why the Approximation Works

Differentiability at \(a\) means that the difference quotient has a limit:

\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=f'(a).

For nonzero \(h\), the quotient can be written as \(f'(a)+\varepsilon(h)\), where \(\varepsilon(h)\to 0\) as \(h\to 0\). Multiplying by \(h\) gives a precise statement of how the function compares with its tangent line.

Rewrite the difference quotient.
For \(h\ne 0\), write \(\frac{f(a+h)-f(a)}{h}=f'(a)+\varepsilon(h)\), where \(\varepsilon(h)\to0\).

Multiply by \(h\) and rearrange.
\(f(a+h)=f(a)+f'(a)h+h\varepsilon(h)\).

Recognize the linearization.
The first two terms are \(L(a+h)=f(a)+f'(a)h\). The remaining difference is \(h\varepsilon(h)\).

Since \(\varepsilon(h)\to0\), the difference between the function and its linearization, divided by \(h\), tends to zero. Equivalently, for \(x\ne a\),

\lim_{x\to a}\frac{f(x)-L(x)}{x-a}=0.

This proves the local approximation claim: compared with the distance \(x-a\), the discrepancy becomes negligible as \(x\) approaches \(a\). It does not say the estimate is exact away from \(a\), nor does it provide a numerical error bound for a particular \(x\). The function must be differentiable at the center, and the estimate is intended for inputs near that center.

Worked Examples

Worked Example: Estimating a Square Root Near 4

Use the linearization of \(f(x)=\sqrt{x}\) at \(a=4\) to estimate \(\sqrt{4.1}\), then check the estimate.

First evaluate the function and derivative at the center:

f(4)=\sqrt{4}=2,\qquad f'(x)=\frac{1}{2\sqrt{x}},\qquad f'(4)=\frac{1}{2\cdot2}=\frac14.

Therefore the linearization is

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

Substitute \(x=4.1\):

L(4.1)=2+\frac14(4.1-4) =2+\frac14(0.1) =2+0.025 =2.025.

Thus \(\sqrt{4.1}\approx 2.025\). To check the direction of the error, square the estimate: \(2.025^2=4.100625\), which is greater than \(4.1\). Both numbers are positive, so \(2.025>\sqrt{4.1}\); the linearization gives a slight overestimate. It also passes through the center, since \(L(4)=2=f(4)\), and its slope is \(1/4=f'(4)\).

Worked Example: Estimating a Reciprocal Near 2

Use the linearization of \(f(x)=1/x\) at \(a=2\) to estimate \(1/2.1\), and compare with the exact value.

Compute the function value and derivative at \(2\):

f(2)=\frac12,\qquad f'(x)=-\frac{1}{x^2},\qquad f'(2)=-\frac14.

The linearization is

L(x)=\frac12-\frac14(x-2).

At \(x=2.1\), the change from the center is \(2.1-2=0.1\), so

L(2.1)=\frac12-\frac14(0.1) =0.5-0.025 =0.475 =\frac{19}{40}.

Thus \(1/2.1\approx0.475\). The exact value is \(1/2.1=10/21\). Their difference is

\frac{10}{21}-\frac{19}{40} =\frac{400-399}{840} =\frac{1}{840}>0.

So the estimate is slightly below the exact value. As a check on the line itself, \(L(2)=1/2=f(2)\), and its slope \(-1/4\) equals \(f'(2)\).

Worked Example: Estimating a Cubic Near 1

Use the linearization of \(f(x)=x^3\) at \(a=1\) to estimate \((1.02)^3\), then compare with the exact value.

At the center,

f(1)=1^3=1,\qquad f'(x)=3x^2,\qquad f'(1)=3.

Hence

$$ L(x)=1+3(x-1). $$

Substitute \(x=1.02\):

$$ L(1.02)=1+3(1.02-1) =1+3(0.02) =1+0.06 =1.06. $$

The estimate is \((1.02)^3\approx1.06\). Direct calculation gives the exact value:

(1.02)^3=\left(\frac{51}{50}\right)^3 =\frac{132651}{125000} =1.061208.

The exact value exceeds the estimate by \(1.061208-1.06=0.001208\). Finally, \(L(1)=1=f(1)\), and the line has slope \(3=f'(1)\), confirming the tangent-line construction.

Choosing a Center and Using the Estimate

To estimate \(f(x)\), choose a center \(a\) that is close to \(x\) and for which \(f(a)\) and \(f'(a)\) are straightforward to find. Then calculate \(x-a\) and use

f(x)\approx f(a)+f'(a)(x-a).

The change \(x-a\) matters. If \(x\) is slightly greater than \(a\), the estimate adjusts \(f(a)\) by the slope times a positive change. If \(x\) is slightly less than \(a\), the change is negative and the adjustment has the opposite sign from the slope. Keeping the signed difference avoids errors when estimating on either side of the center.

A useful check is to evaluate the proposed line at the center. It must give \(L(a)=f(a)\). Also check that the coefficient of \(x-a\) is \(f'(a)\), not a derivative evaluated at the target input. These checks catch common setup errors before an estimate is made.

A Common Mistake

A frequent error is to use \(f(a)+f'(a)x\), forgetting to subtract the center inside the slope term. For \(f(x)=x^3\) at \(a=2\), this incorrect expression would be

$$ 8+12x. $$

At \(x=2\), it gives \(8+12(2)=32\), even though the function value is \(f(2)=8\). Therefore it cannot be the tangent line at \((2,8)\).

The correct construction uses \(x-a\): \(f(2)=8\) and \(f'(2)=3(2)^2=12\), so

$$ L(x)=8+12(x-2). $$

Now \(L(2)=8+12(0)=8\), and the slope is \(12\), as required. For example, at \(x=2.1\), this line gives \(L(2.1)=8+12(0.1)=9.2\). The center subtraction is essential: it makes the change in the input, rather than the input itself, control the adjustment to \(f(a)\).

Why Linearization Matters

Linearization replaces a local question about a possibly nonlinear function with an easy calculation involving a value and a slope. This is useful for mental estimates, checking numerical calculations, and understanding how small input changes affect output. In applications, the derivative often describes a rate of change, so \(f'(a)(x-a)\) estimates the corresponding change in the function near \(a\).

Accuracy is not automatic. Differentiability guarantees that the tangent line is a good local approximation in the precise limiting sense derived above, but it does not say how close a particular estimate will be. The examples show why comparison with an exact value, when one is available, is a valuable check. For other functions, use the model with a nearby center and interpret the result as an estimate rather than an identity.

Check Your Understanding

For each question, use the center and function given, and show how the value and derivative determine the linearization.

Find the linearization of \(f(x)=x^2\) at \(a=3\).
Use the linearization of \(f(x)=1/x\) at \(a=1\) to estimate \(1/1.05\).
Why must a linearization at \(a\) pass through \((a,f(a))\)?
What does differentiability at \(a\) tell us about the difference between \(f(x)\) and its linearization as \(x\) approaches \(a\)?
Identify the setup error in \(L(x)=f(a)+f'(a)x\), and write the correct general formula.