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].
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].
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.
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.
Hint
[latex]L(x)=f(a)+f^{\prime}(a)(x-a)[/latex]
Show Solution
[latex]L(x)=2+\frac{1}{12}(x-8)[/latex]; 2.00833
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].
Show Solution
First we note that since [latex]\frac{\pi}{3}[/latex] rad is equivalent to [latex]60^{\circ}[/latex], using the linear approximation at [latex]x=\pi /3[/latex] seems reasonable. The linear approximation is given by
[latex]L(x)=f(\frac{\pi}{3})+f^{\prime}(\frac{\pi}{3})(x-\frac{\pi}{3})[/latex].
We see that
[latex]\begin{array}{lll}f(x)= \sin x & \Rightarrow & f(\frac{\pi}{3})= \sin (\frac{\pi}{3})=\frac{\sqrt{3}}{2} \\ f^{\prime}(x)= \cos x & \Rightarrow & f^{\prime}(\frac{\pi}{3})= \cos (\frac{\pi}{3})=\frac{1}{2} \end{array}[/latex]
Therefore, the linear approximation of [latex]f[/latex] at [latex]x=\pi /3[/latex] is given by Figure 3.
[latex]L(x)=\frac{\sqrt{3}}{2}+\frac{1}{2}(x-\frac{\pi}{3})[/latex]
To estimate [latex] \sin (62^{\circ})[/latex] using [latex]L[/latex], we must first convert [latex]62^{\circ}[/latex] to radians. We have [latex]62^{\circ}=\frac{62\pi}{180}[/latex] radians, so the estimate for [latex] \sin (62^{\circ})[/latex] is given by
[latex] \sin (62^{\circ})=f(\frac{62\pi}{180})\approx L(\frac{62\pi }{180})=\frac{\sqrt{3}}{2}+\frac{1}{2}(\frac{62\pi }{180}-\frac{\pi }{3})=\frac{\sqrt{3}}{2}+\frac{1}{2}(\frac{2\pi }{180})=\frac{\sqrt{3}}{2}+\frac{\pi }{180}\approx 0.88348[/latex].
Figure 3. The linear approximation to [latex]f(x)= \sin x[/latex] at [latex]x=\frac{\pi}{3}[/latex] provides an approximation to [latex] \sin x[/latex] for [latex]x[/latex] near [latex]\frac{\pi}{3}.[/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].
Hint
[latex]L(x)=f(a)+f^{\prime}(a)(x-a)[/latex]
Show Solution
[latex]L(x)=−x+\frac{\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].
Show Solution
The linear approximation at [latex]x=0[/latex] is given by
[latex]L(x)=f(0)+f^{\prime}(0)(x-0)[/latex].
Because
[latex]\begin{array}{lll} f(x)=(1+x)^n & \Rightarrow & f(0)=1 \\ f^{\prime}(x)=n(1+x)^{n-1} & \Rightarrow & f^{\prime}(0)=n, \end{array}[/latex]
the linear approximation is given by Figure 4a.
[latex]L(x)=1+n(x-0)=1+nx[/latex]
We can approximate [latex](1.01)^3[/latex] by evaluating [latex]L(0.01)[/latex] when [latex]n=3[/latex]. We conclude that
[latex](1.01)^3=f(1.01)\approx L(1.01)=1+3(0.01)=1.03[/latex].
Figure 4. (a) The linear approximation of [latex]f(x)[/latex] at [latex]x=0[/latex] is [latex]L(x)[/latex]. (b) The actual value of [latex]1.01^3[/latex] is 1.030301. The linear approximation of [latex]f(x)[/latex] at [latex]x=0[/latex] estimates [latex]1.01^3[/latex] to be 1.03.
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.
Hint
[latex]f^{\prime}(x)=4(1+x)^3[/latex]