Linear Approximation of a Function at a Point

Learning Outcomes

  • Describe the linear approximation to a function at a point.
  • Write the linearization of a given function.

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

[latex]y=f(a)+f^{\prime}(a)(x-a)[/latex]

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

[latex]y=\dfrac{1}{2}-\dfrac{1}{4}(x-2)[/latex]

 

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

[latex]y=\dfrac{1}{2}-\dfrac{1}{4}(2.1-2)=0.475[/latex]

 

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

[latex]f(2.1)=\dfrac{1}{2.1}\approx 0.47619[/latex]

 

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

[latex]y=\frac{1}{2}-\dfrac{1}{4}(10-2)=\dfrac{1}{2}-2=-1.5[/latex],

 

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

This figure has two parts a and b. In figure a, the line f(x) = 1/x is shown with its tangent line at x = 2. In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near x = 2.

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

[latex]f(x)\approx f(a)+f^{\prime}(a)(x-a)[/latex] for [latex]x[/latex] near [latex]a[/latex]

We call the linear function

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

 

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].

Watch the following video to see the worked solution to Example: Linear Approximation of [latex]\sqrt{x}[/latex].

Try It

Find the local linear approximation to [latex]f(x)=\sqrt[3]{x}[/latex] at [latex]x=8[/latex]. Use it to approximate [latex]\sqrt[3]{8.1}[/latex] to five decimal places.

Example: Linear Approximation of [latex] \sin x[/latex]

Find the linear approximation of [latex]f(x)= \sin x[/latex] at [latex]x=\dfrac{\pi}{3}[/latex] and use it to approximate [latex]\sin (62^{\circ})[/latex].

Watch the following video to see the worked solution to Example: Linear Approximation of [latex] \sin x[/latex].

Try It

Find the linear approximation for [latex]f(x)= \cos x[/latex] at [latex]x=\dfrac{\pi }{2}[/latex].

Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for [latex]f(x)=(1+x)^n[/latex] at [latex]x=0[/latex], 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 [latex]f(x)=(m+x)^n[/latex] to estimate roots and powers near a different number [latex]m[/latex].

Example: Approximating Roots and Powers

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

Try It

Find the linear approximation of [latex]f(x)=(1+x)^4[/latex] at [latex]x=0[/latex] without using the result from the preceding example.

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Try It