Consider a function [latex]f[/latex] that is differentiable at a point [latex]x=a[/latex]. Recall that the tangent line to the graph of [latex]f[/latex] at [latex]a[/latex] is given by the equation

This is simply derived from the point-slope form of the equation of a line [latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex] by adding  [latex]{y}_{1}[/latex] to both sides!

For example, consider the function [latex]f(x)=\frac{1}{x}[/latex] at [latex]a=2[/latex]. Since [latex]f[/latex] is differentiable at [latex]x=2[/latex] and [latex]f^{\prime}(x)=-\frac{1}{x^2}[/latex], we see that [latex]f^{\prime}(2)=-\frac{1}{4}[/latex]. Therefore, the tangent line to the graph of [latex]f[/latex] at [latex]a=2[/latex] is given by the equation

Figure 1a shows a graph of [latex]f(x)=\frac{1}{x}[/latex] along with the tangent line to [latex]f[/latex] at [latex]x=2[/latex]. Note that for [latex]x[/latex] near 2, the graph of the tangent line is close to the graph of [latex]f[/latex]. As a result, we can use the equation of the tangent line to approximate [latex]f(x)[/latex] for [latex]x[/latex] near 2. For example, if [latex]x=2.1[/latex], the [latex]y[/latex] value of the corresponding point on the tangent line is

The actual value of [latex]f(2.1)[/latex] is given by

Therefore, the tangent line gives us a fairly good approximation of [latex]f(2.1)[/latex] (Figure 1b). However, note that for values of [latex]x[/latex] far from 2, the equation of the tangent line does not give us a good approximation. For example, if [latex]x=10[/latex], the [latex]y[/latex]-value of the corresponding point on the tangent line is

whereas the value of the function at [latex]x=10[/latex] is [latex]f(10)=0.1[/latex].

Figure 1. (a) The tangent line to [latex]f(x)=\frac{1}{x}[/latex] at [latex]x=2[/latex] provides a good approximation to [latex]f[/latex] for [latex]x[/latex] near 2. (b) At [latex]x=2.1[/latex], the value of [latex]y[/latex] on the tangent line to [latex]f(x)=\frac{1}{x}[/latex] is 0.475. The actual value of [latex]f(2.1)[/latex] is [latex]\frac{1}{2.1}[/latex], which is approximately 0.47619.

In general, for a differentiable function [latex]f[/latex], the equation of the tangent line to [latex]f[/latex] at [latex]x=a[/latex] can be used to approximate [latex]f(x)[/latex] for [latex]x[/latex] near [latex]a[/latex]. Therefore, we can write

We call the linear function

the linear approximation, or tangent line approximation, of [latex]f[/latex] at [latex]x=a[/latex]. This function [latex]L[/latex] is also known as the linearization of [latex]f[/latex] at [latex]x=a[/latex].

To show how useful the linear approximation can be, we look at how to find the linear approximation for [latex]f(x)=\sqrt{x}[/latex] at [latex]x=9[/latex].

Example: Linear Approximation of [latex]\sqrt{x}[/latex]

Find the linear approximation of [latex]f(x)=\sqrt{x}[/latex] at [latex]x=9[/latex] and use the approximation to estimate [latex]\sqrt{9.1}[/latex].

Show Solution

Since we are looking for the linear approximation at [latex]x=9[/latex], using the tangent line approximation, we know the linear approximation is given by

[latex]L(x)=f(9)+f^{\prime}(9)(x-9)[/latex].

 

We need to find [latex]f(9)[/latex] and [latex]f^{\prime}(9)[/latex].

[latex]\begin{array}{lll} f(x)=\sqrt{x}& \Rightarrow & f(9)=\sqrt{9}=3 \\ f^{\prime}(x)=\frac{1}{2\sqrt{x}}& \Rightarrow & f^{\prime}(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6} \end{array}[/latex]

 

Therefore, the linear approximation is given by Figure 2.

[latex]L(x)=3+\frac{1}{6}(x-9)[/latex]

 

Using the linear approximation, we can estimate [latex]\sqrt{9.1}[/latex] by writing

[latex]\sqrt{9.1}=f(9.1)\approx L(9.1)=3+\frac{1}{6}(9.1-9)\approx 3.0167[/latex].

Figure 2. The local linear approximation to [latex]f(x)=\sqrt{x}[/latex] at [latex]x=9[/latex] provides an approximation to [latex]f[/latex] for [latex]x[/latex] near 9.

Analysis

Using a calculator, the value of [latex]\sqrt{9.1}[/latex] to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate [latex]\sqrt{x}[/latex], at least for [latex]x[/latex] near 9. At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate [latex]\sqrt{9.1}[/latex]. However, how does the calculator evaluate [latex]\sqrt{9.1}[/latex]? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.