The Linearization Formula at a Base Point

What the Linearization Formula Captures

A differentiable function may have a complicated graph, but at a particular input its value and derivative give a simpler description: a line that passes through the corresponding point on the graph and has the same slope there. This line is called the function’s linearization at that input.

The formula is compact, but each part has a specific role. The input \(a\) is the base point, \(f(a)\) gives the height of the line there, and \(f'(a)\) gives its slope. Keeping those roles clear makes the formula straightforward to construct and check.

Definition: Suppose \(f\) is defined on an open interval containing \(a\), and \(f\) is differentiable at \(a\). The linearization of \(f\) at the base point \(a\) is the function \(L(x)=f(a)+f'(a)(x-a)\). It is the line through \((a,f(a))\) with slope \(f'(a)\).

Reading the Formula

In \(L(x)=f(a)+f'(a)(x-a)\), the variable \(x\) is the input to the linearization. The number \(a\) is fixed: it is the base point at which the function’s value and derivative are evaluated. The expression \(x-a\) measures the horizontal displacement from that point.

Base point: \(a\), the input where the function and its derivative are evaluated.
Value at the base point: \(f(a)\), the vertical coordinate of the point on the original graph.
Slope: \(f'(a)\), the slope of the linearization.
Input displacement: \(x-a\), the change in input from the base point.

The formula is the point-slope equation of a line. In that equation, the point is \((a,f(a))\), the slope is \(f'(a)\), and the input is \(x\). The linearization is therefore determined by two pieces of information about \(f\) at \(a\): its value and its derivative.

Although the name contains “linear,” \(L\) is generally an affine function: it has the form \(mx+b\), possibly with a nonzero constant term. In calculus, “linearization” refers to this first-degree model built at a point.

Why the Formula Has This Form

A line with slope \(m\) passing through the point \((a,y_0)\) satisfies the point-slope equation \(y-y_0=m(x-a)\). To construct a line that matches \(f\) at \(a\), its point must be \((a,f(a))\). To match the slope of \(f\) there, its slope must be \(f'(a)\). Substitution into the point-slope equation gives the linearization formula.

$$ y-f(a)=f'(a)(x-a) $$

$$ y=f(a)+f'(a)(x-a) $$

This construction also verifies the two defining properties directly. First, set \(x=a\). Then \(x-a=0\), so the line has the same value as \(f\) at the base point. Second, the derivative of the line is its constant slope, which is \(f'(a)\).

$$ L(a)=f(a)+f'(a)(a-a)=f(a) $$

L'(x)=f'(a) \qquad\text{and therefore}\qquad L'(a)=f'(a)

Thus the linearization matches both the function’s value and its slope at \(a\). This is the precise sense in which the line represents the function locally at its base point.

Constructing a Linearization Step by Step

For a given function and base point, the calculation follows a consistent order. Evaluate the function at the base point, find the derivative of the function, evaluate that derivative at the same base point, and substitute both results into the formula. Simplify only after the formula is assembled.

Record the base point.
Identify the specified input \(a\). Use this same value in both \(f(a)\) and \(f'(a)\).

Find the function value.
Calculate \(f(a)\), which supplies the point’s vertical coordinate.

Find the slope.
Differentiate \(f\), then evaluate the derivative at the base point to obtain \(f'(a)\).

Substitute and check.
Write \(L(x)=f(a)+f'(a)(x-a)\), then verify \(L(a)=f(a)\) and \(L'(a)=f'(a)\).

Worked Examples

Worked Example: A Polynomial at \(a=2\)

Find the linearization of \(f(x)=x^2+3x-1\) at the base point \(a=2\).

First evaluate the function:

$$ f(2)=2^2+3(2)-1=4+6-1=9 $$

Differentiate and evaluate the derivative at the same point:

$$ f'(x)=2x+3 $$

$$ f'(2)=2(2)+3=4+3=7 $$

Substitute \(a=2\), \(f(2)=9\), and \(f'(2)=7\) into the formula:

$$ L(x)=9+7(x-2)=9+7x-14=7x-5 $$

Check the value at the base point: \(L(2)=7(2)-5=14-5=9=f(2)\). The derivative is \(L'(x)=7\), so \(L'(2)=7=f'(2)\). The constructed line has the required point and slope.

Worked Example: A Square Root at \(a=4\)

Find the linearization of \(f(x)=\sqrt{x}\) at \(a=4\). The function is defined and differentiable on an open interval around \(4\).

Evaluate the function:

f(4)=\sqrt{4}=2

Differentiate, then evaluate at \(4\):

f'(x)=\frac{1}{2\sqrt{x}}

f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{2\cdot2}=\frac14

Insert these values into the linearization formula and simplify:

L(x)=2+\frac14(x-4) =2+\frac{x}{4}-1 =1+\frac{x}{4}

Check the value: \(L(4)=1+\frac44=2=f(4)\). Also, \(L'(x)=\frac14\), so \(L'(4)=\frac14=f'(4)\). Both defining properties hold.

Worked Example: Sine at \(a=\frac{\pi}{6}\)

Find the linearization of \(f(x)=\sin x\) at \(a=\frac{\pi}{6}\), with angles measured in radians.

Use the exact value of sine at the base point:

f\left(\frac{\pi}{6}\right)=\sin\left(\frac{\pi}{6}\right)=\frac12

The derivative of \(\sin x\) is \(\cos x\). Its value at the base point is:

f'(x)=\cos x

f'\left(\frac{\pi}{6}\right) =\cos\left(\frac{\pi}{6}\right) =\frac{\sqrt{3}}{2}

Substitute the base point, function value, and slope:

L(x)=\frac12+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)

At the base point, the displacement is zero, so \(L\left(\frac{\pi}{6}\right)=\frac12=f\left(\frac{\pi}{6}\right)\). Differentiating the line gives \(L'(x)=\frac{\sqrt{3}}{2}\), hence \(L'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}=f'\left(\frac{\pi}{6}\right)\). The formula has the correct value and slope.

Why the Base Point Must Stay Visible

The expression \(x-a\) is not an optional detail. It measures how far the input is from the point where the function value and slope were found. At \(x=a\), that displacement becomes zero, so the linearization automatically passes through the correct point.

For example, if \(f(x)=x^2\) and \(a=3\), then \(f(3)=9\) and \(f'(x)=2x\), giving \(f'(3)=6\). The correct formula is \(L(x)=9+6(x-3)\). If the displacement is mistakenly replaced by \(x\), the incorrect expression becomes \(9+6x\). At the base point it gives \(9+6(3)=27\), not \(f(3)=9\). The correction is to include \(x-a\): \(9+6(x-3)=6x-9\). Now \(L(3)=18-9=9\), and \(L'(x)=6\), as required.

Common pitfall: Do not use \(f'(x)\) in the linearization formula where the fixed slope \(f'(a)\) belongs. The linearization is a line, so its slope is constant. Also, do not omit the displacement \(x-a\); omitting it usually makes the line miss the required point.

What the Formula Does—and Does Not—Assert

The formula defines a line from information about \(f\) at one point. It guarantees that the line agrees with the function in value and derivative at the base point. It does not say that the function and the line have equal values at every input, or that \(f\) itself is a line.

The formula also requires the derivative \(f'(a)\) to exist. If \(f\) is not differentiable at \(a\), this definition does not provide a linearization there. When \(f\) is differentiable at \(a\), the base-point value and slope give a precise, single line, even if the formula is written in different but algebraically equivalent forms.

Check Your Understanding

For each question, identify which quantities are fixed by the base point and use the defining properties to check your reasoning.

In \(L(x)=f(a)+f'(a)(x-a)\), which quantity gives the slope of the linearization?
Suppose \(f(1)=4\) and \(f'(1)=-3\). Write the linearization at \(a=1\), and verify its value at the base point.
For \(f(x)=x^2+1\) at \(a=2\), find \(f(a)\), \(f'(a)\), and the linearization.
Why does the formula use \(x-a\) rather than just \(x\)?
What condition on \(f\) at the base point is required for the linearization to be defined?

The Linearization Formula at a Base Point

What You'll Learn

What the Linearization Formula Captures

Reading the Formula

Why the Formula Has This Form

Constructing a Linearization Step by Step