4.2 Linear approximations and differentials

We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions .

Linear approximation of a function at a point

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

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

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

The actual value of $f(2.1)$ is given by

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

whereas the value of the function at $x=10$ is $f(10)=0.1.$

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

We call the linear function

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

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

Linear approximation of $\sqrt{x}$

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

Since we are looking for the linear approximation at $x=9,$ using [link] we know the linear approximation is given by

$L(x)=f(9)+f\prime (9)(x-9).$

We need to find $f(9)$ and $f\prime (9).$

$\begin{array}{ccc}\hfill f(x)=\sqrt{x}& \Rightarrow \hfill & f(9)=\sqrt{9}=3\hfill \\ \hfill f\prime (x)=\frac{1}{2\sqrt{x}}& \Rightarrow \hfill & f\prime (9)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\hfill \end{array}$

Therefore, the linear approximation is given by [link] .

$L(x)=3+\frac{1}{6}(x-9)$

Using the linear approximation, we can estimate $\sqrt{9.1}$ by writing

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

Got questions? Get instant answers now!

Got questions? Get instant answers now!