OpenStax Calculus: Volume 1

Subsection 4.2.1 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 \(f'(x)=-\frac{1}{x^2},\) we see that \(f'(2)=-\frac{1}{4}.\) Therefore, the tangent line to the graph of \(f\) at \(a=2\) is given by the equation

Figure 4.11(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 \(f(2.1)\) (Figure 4.11(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 \(x=9.\)

Example 4.12.

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

Solution.

Since we are looking for the linear approximation at \(x=9,\) using (4.2.1) we know the linear approximation is given by

\begin{equation*} L(x)=f(9)+f'(9)(x-9). \end{equation*}

We need to find \(f(9)\) and \(f'(9).\)

\begin{align*} f(x)=\sqrt{x}\amp \implies \amp \amp f(9)=\sqrt{9}=3\\ f'(x)=\frac{1}{2\sqrt{x}}\amp\implies \amp \amp f'(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6} \end{align*}

Therefore, the linear approximation is given by Figure 4.13.

\begin{equation*} L(x)=3+\frac{1}{6}(x-9) \end{equation*}

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

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

Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for \(f(x)=(1+x)^n\) at \(x=0,\) which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form \(f(x)=(m+x)^n \) to estimate roots and powers near a different number \(m.\)

Example 4.16. Approximating Roots and Powers.

Find the linear approximation of \(f(x)=(1+x)^n\) at \(x=0.\) Use this approximation to estimate \((1.01)^3.\)

Solution.

The linear approximation at \(x=0\) is given by

\begin{equation*} L(x)=f(0)+f'(0)(x-0). \end{equation*}

Because

\begin{align*} f(x)=(1+x)^n\amp\implies \amp \amp f(0)=1\\ f'(x)=n(1+x)^{n-1}\amp\implies \amp \amp f'(0)=n, \end{align*}

the linear approximation is given by Figure 4.17(a).

\begin{equation*} L(x)=1+n(x-0)=1+nx \end{equation*}

We can approximate \((1.01)^3\) by evaluating \(L(0.01)\) when \(n=3.\) We conclude that

\begin{equation*} (1.01)^3=f(1.01)\approx L(1.01)=1+3(0.01)=1.03. \end{equation*}

Subsection 4.2.2 Average Cost

Subsection 4.2.3 Key Concepts

• A differentiable function \(y=f(x)\) can be approximated at \(a\) by the linear function

  \begin{equation*} L(x)=f(a)+f'(a)(x-a). \end{equation*}

Subsection 4.2.4 Key Equations

• Linear approximation \(L(x)=f(a)+f'(a)(x-a)\)

• Average Cost Function \(\overline{C}(x)=\frac{C(x)}{x}\)