## Linearization of a Function

Recall the definition of $$f'(x_{0})$$: $\lim_{\Delta x\to0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}=f'(x_{0}).$ From the definition of a limit, we know that for $$x$$ close to $$x_{0}$$, the fraction $$[f(x_{0}+\Delta x)-f(x_{0})]/\Delta x$$ must be close to $$f'(x_{0})$$; that is, for $$x$$ close to $$x_{0}$$ $\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}\approx f'(x_{0}).$ Multiplying both sides by $$\Delta x$$, we get $f(x_{0}+\Delta x)-f(x_{0})\approx f'(x_{0})\Delta x.$ In terms of $$x=x_{0}+\Delta x$$, we have $f(x)-f(x_{0})\approx f'(x_{0})(x-x_{0})$ or $f(x)\approx f(x_{0})+f'(x_{0})(x-x_{0}).$ Let’s denote the right-hand side by $$L(x)$$: $L(x)=f(x_{0})+f'(x_{0})(x-x_{0}).$ The graph of the function $$L$$ is the tangent line to the curve $$y=f(x)$$ at $$\left(x_{0},f(x_{0})\right)$$.1 It is also clear from Figure 1 that the tangent line at $$\left(x_{0},f(x_{0})\right)$$ is a good approximation to $$f(x)$$ when $$x$$ is close to $$x_{0}$$.

Definition. The function $$L$$ $L(x)=f(x_{0})+f'(x_{0})(x-x_{0})$ is called the linearization of $$f$$ at $$x_{0}$$, and the approximation $$f(x)\approx L(x)$$ or $f(x)\approx f(x_{0})+f'(x_{0})(x-x_{0}).$ is called the linear approximation or tangent line approximation of $$f$$ at $$x_{0}$$.2
• If we use the delta notation $$\Delta y=f(x)-f(x_{0})$$ and $$\Delta x=x-x_{0}$$, then we can write the linear approximation as $\underbrace{\Delta y}_{f(x)-f(x_{0})}\approx f'(x)\underbrace{\Delta x}_{x-x_{0}}$ Figure 1. The tangent line to the curve $$y=f(x)$$ at $$(x_{0},f(x_{0}))$$ is the line $$L(x)=f(x_{0})+f'(x_{0})(x-x_{0})$$. When $$x$$ is close to $$x_{0}$$, $$f(x)\approx L(x)$$.

For example, if $$y=f(x)=x^{2}$$, then $f'(x)=2x$ and the linear approximation is $\underbrace{f(x+\Delta x)}_{(x+\Delta x)^{2}}\approx\underbrace{f(x)}_{x^{2}}+2x\Delta x$ and by expanding $$(x+\Delta x)^{2}$$, we have $x^{2}+2x\,\Delta x+\left(\Delta x\right)^{2}\approx x^{2}+2x\,\Delta x.$ This approximation tells us that if $$\Delta x$$ is small, $$(\Delta x)^{2}$$ is negligible. This makes sense, because when $$\Delta x$$ is small, $$(\Delta x)^{2}$$ is the square of a small number and this is much smaller than $$\Delta x$$ itself. For example, if $$\Delta x=0.001$$ then $$(\Delta x)^{2}=0.000000001$$.

For the geometrical interpretation of this approximation, consider a square having sides $$x$$ by $$x$$, and suppose that each side is increased to $$x+\Delta x$$ (Figure 2). The enlarged square is made up of the original square of area $$x^{2}$$, two rectangles at the top and on the right, each of which is of area $$x\cdot\Delta x$$, and the little square at the top right corner of area $$(\Delta x)^{2}$$. We may ignore this little square and say that the area of the enlarged square is approximately $$x^{2}+2x\,\Delta x.$$

Example 1

Use the linearization of $$f(x)=\sqrt{x}$$ at $$x_{0}=9$$ to approximate $$\sqrt{10}$$ and $$\sqrt{8}$$.

Solution 1

The linearization of $$f$$ at $$x=9$$ is $L(x)=f(9)+f'(9)(x-9).$ Because $f(x)=\sqrt{x}\Rightarrow f'(x)=\frac{1}{2\sqrt{x}},$ we have $$f(9)=3$$ and $$f'(9)=1/6$$, and thus $L(x)=3+\frac{1}{6}(x-9).$ The graphs of $$f$$ and $$L$$ are illustrated in Figure 3.

At $$x=10$$, the linearization gives $L(10)=3+\frac{1}{6}\cdot(10-9)\approx3.16667.$ The true value of $$\sqrt{10}$$ to 5 digits is 3.16228.

At $$x=8$$, the linearization gives $L(8)=3+\frac{1}{6}\cdot(8-9)\approx2.83333.$ The true value of $$\sqrt{8}$$ to 5 digits is 2.8284. Figure 3. The graghs of $$f(x)=\sqrt{3}$$ and its linearization at $$x=9$$, $$L(x)=3+(x-9)/6$$.
Example 2

Approximate $$\sin(31^{\circ})$$ and $$\sin(29^{\circ})$$.

Solution 2

Because we know the exact value of $$\sin(30^{\circ})=\sin(\pi/6)$$, we can use the linearization of $$f(x)=\sin x$$ at $$x=\pi/6$$ to approximate $$\sin(31^{\circ})$$ and $$\sin(29^{\circ})$$. The derivative of $$f$$ is $f'(x)=\cos x.$ So we have $$f(\pi/6)=1/2$$ and $$f'(\pi/6)=\cos(\pi/6)=\sqrt{3}/2$$.

The linearization of $$f$$ at $$x=\pi/6$$ is $L(x)=\underbrace{\frac{1}{2}}_{f(\pi/6)}+\underbrace{\frac{\sqrt{3}}{2}}_{f'(\pi/6)}\underbrace{\Delta x}_{\left(x-\frac{\pi}{6}\right)}.$ Notice that $$\frac{d}{dx}\sin x=\cos x$$ works only when $$x$$ is in radian measure. Therefore, $$\Delta x$$ is also in radians.

Because $1^{\circ}=\frac{\pi}{180}\text{ radian,}$ we have

\begin{aligned} \sin(31^{\circ}) & \approx L\left(\frac{\pi}{6}+\frac{\pi}{180}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(\frac{\pi}{180}\right)\\ & \approx0.5151.\end{aligned} and \begin{aligned} \sin(29^{\circ}) & \approx L\left(\frac{\pi}{6}-\frac{\pi}{180}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(-\frac{\pi}{180}\right)\\ & \approx0.4849.\end{aligned} The true values of $$\sin(31^{\circ})$$ and $$\sin(29^{\circ})$$ to 4 digits are 0.5150 and 0.4848, respectively. The graphs of $$f$$ and its linearization are illustrated in Figure 4. Figure 4. The graphs of $$y=\sin x$$ and $$y=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)$$.
Example 3

What is the linear approximation of $$f(x)=\sin x$$ at $$x=0$$? For what values of $$x$$ is this linear approximation accurate to within 0.01?

Solution 3

Since $f'(x)=\cos x,$ we have $$f(0)=\sin0=0$$ and $$f'(0)=\cos0=1$$. Therefore, the linearization of $$f$$ is \begin{aligned} L(x) & =f(0)+f'(0)(x-0)\\ & =0+1(x-0)=x.\end{aligned} The graphs of $$f$$ and $$L$$ are shown in Figure 5. Figure 5. The graph of $$y=f(x)=\sin x$$ and its linear approximation at $$x=0$$.

The accuracy to within 0.01 means that the difference between $$f$$ and $$L$$ is less than 0.01; that is, $|f(x)-L(x)|<0.01$ or $|\sin x-x|<0.01.$ The graph of $$y=|\sin x-x|$$ is shown in Figure 6 (the blue curve). In this figure, the vertical lines show where $$y=|\sin x-x|$$ meets the horizontal line $$y=0.01$$. If $$x$$ is between these two vertical lines, the error (i.e. $$|\sin x-x|$$) is less than $$0.01$$. The vertical lines are at $$x\approx0.39$$ and $$x\approx-0.39$$. Notice that here $$x$$ is radian measure. So if $-0.39\text{ rad}<x<0.39\text{ rad}$ or $-22.4^{\circ}<x<22.4^{\circ},$ the difference between $$\sin x$$ and $$x$$ is less than 0.01. Figure 6. When $$-0.39 ## Is the Linear Approximation Always a Good Approximation? #### Show the answer The linear approximation is based on the assumption that the tangent line remains close to the curve \(y=f(x)$$ when $$x$$ is close to $$x_{0}$$. If the curve has a pronounced bend at $$\left(x_{0},f(x_{0})\right)$$ (for example, see Figure 7), then the linear approximation does not do a good job unless $$x$$ is really close to $$x_{0}$$. The bending of a curve is measured by the second derivative. We will talk about the bending of curves in the next chapter.  Figure 7. The linear approximation does not do a good job if the curve is bending sharply near $$(x_{0},f(x_{0}))$$ unless $$x$$ is really close to $$x_{0}$$.

## An Application of Linearization in Physics Problems

#### Show the application

Linearization can simplify some equations. Here is an example:

A simple pendulum is a body of mass $$m$$ on the end of a massless string suspended from a point (Figure 8). When displaced a little bit and released, the pendulum swings back and forth. Let $$\theta$$ denote the angle that the string makes with the vertical. The displacement of the body from equilibrium is then $$s=L\theta$$, where $$L$$ is the length of the string. The force along the direction of motion is $$-mg\sin\theta$$ where $$g$$ is acceleration due to gravity ($$g\approx9.81$$ m/s$$^{2}$$ or $$g\approx32$$ ft/s$$^{2}$$). It follows from Newton’s second law that $F=ma=-mg\sin\theta\Rightarrow a=-g\sin\theta.$ Because $$a=d^{2}s/dt^{2}$$ and $$s=L\theta$$ we have $\underbrace{\frac{d^{2}}{dt^{2}}(L\theta)}_{a}=-g\sin\theta$ or $L\frac{d^{2}\theta}{dt^{2}}=-g\sin\theta.$ To describe the oscillation of the pendulum, we look for a function $$\theta=f(t)$$ that satisfies the following equation: $\underbrace{f”(t)}_{\frac{d^{2}\theta}{dt^{2}}}+\frac{g}{L}\sin(\underbrace{f(t)}_{\theta})=0.$ There is no function (among the functions that we know) that can satisfy the above equation. However, we can use the linear approximation $$\sin\theta\approx\theta$$ or $$\sin(f(t))\approx f(t)$$ and rewrite the equation as $f”(t)+\frac{g}{L}f(t)=0.$ We can show that the solution of this equation is $f(t)=\theta_{0}\cos\left(\sqrt{\frac{g}{L}}\ t\right),$ where $$\theta_{0}=f(0)$$ is initial angular displacement. As long as $$\theta_{0}$$ is small (about $$22.5^{\circ}$$ or less as we showed in the last example), the linear approximation is valid.

1. Recall that the equation of a line through $$(x_{0},y_{0})$$ with slope $$m$$ is $$y-y_{0}=m(x-x_{0})$$ or $$y=y_{0}+m(x-x_{0}).$$ So because the slope of the tangent line to the curve $$y=f(x)$$ at $$\left(x_{0},f(x_{0})\right)$$ is $$f'(x_{0})$$, its equation is $$y=f(x_{0})+f'(x_{0})(x-x_{0}).$$↩︎
2. Notice that $$L$$ is not a linear function unless $$f(x_{0})=0$$. However, even if $$f(x_{0})\neq0$$, we may loosely call $$L$$ a linear function because its graph is a line.↩︎