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# 2.7: Linearization and Newton’s Method

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Approximate a function by the method of linearization.
• Know Newton’s Method for approximating roots of a function.

## Linearization: The Tangent Line Approximation

If f\begin{align*}f\end{align*} is a differentiable function at x0\begin{align*}x_0\end{align*}, then the tangent line, y=mx+b\begin{align*}y = mx + b\end{align*}, to the curve y=f(x)\begin{align*}y = f(x)\end{align*} at x0\begin{align*}x_0\end{align*} is a good approximation to the curve y=f(x)\begin{align*}y = f(x)\end{align*} for values of x\begin{align*}x\end{align*} near x0\begin{align*}x_0\end{align*} (Figure 8a). If you “zoom in” on the two graphs, y=f(x)\begin{align*}y = f(x)\end{align*} and the tangent line, at the point of tangency, \begin{align*}(x_0, f(x_0))\end{align*}, or if you look at a table of values near the point of tangency, you will notice that the values are very close (Figure 8b).

Since the tangent line passes through point \begin{align*}(x_0, f(x_0))\end{align*} and the slope is \begin{align*}f'(x_0)\end{align*}, we can write the equation of the tangent line, in point-slope form, as

\begin{align*}y - y_0 &= m(x - x_0)\\ y - f(x_0) &= f'(x_0)(x - x_0)\end{align*}

Solving for \begin{align*}y\end{align*},

\begin{align*}y = f(x_0) + f'(x_0) (x - x_0)\end{align*}

Figure 8a

Figure 8b

So for values of \begin{align*}x\end{align*} close to \begin{align*}x_0\end{align*}, the values of \begin{align*}y\end{align*} of this tangent line will closely approximate \begin{align*}f(x)\end{align*}. This gives the approximation

\begin{align*}f(x) = f(x_0) + f'(x_0) (x - x_0).\end{align*}

The Tangent Line Approximation (Linearization)

If \begin{align*}f\end{align*} is a differentiable function at \begin{align*}x = x_0\end{align*}, then the approximation function

\begin{align*} L(x) = f(x) \approx f(x_0) + f'(x_0) (x - x_0)\end{align*}

is a linearization of \begin{align*}f\end{align*} near \begin{align*}x_0\end{align*}.

Example 1:

Find the linearization of \begin{align*} f(x) = \sqrt{x + 3}\end{align*} at point \begin{align*}x = 1\end{align*}.

Solution:

Taking the derivative of \begin{align*}f(x)\end{align*},

\begin{align*} f'(x) & = \frac{1} {2} (x + 3)^{-1/2},\end{align*}

we have \begin{align*}f(1) = \sqrt{4} = 2, f'(1) = 1/4,\end{align*} and

\begin{align*}f(x) & \approx f(x_0) + f'(x_0)(x - x_0)\\ & \approx 2 + \frac{1} {4} (x - 1) \\ & \approx \frac{1} {4}x + \frac{7} {4}.\end{align*}

This tells us that near the point \begin{align*}x = 1\end{align*}, the function \begin{align*} f(x) = \sqrt{x + 3}\end{align*} approximates the line \begin{align*}y = (x/4) + 7/4\end{align*}. As we move away from \begin{align*}x = 1\end{align*}, we lose accuracy (Figure 9).

Figure 9

Example 2:

Find the linearization of \begin{align*}y = \sin x\end{align*} at \begin{align*}x = \pi/3\end{align*}.

Solution:

Since \begin{align*}f(\pi/3) = \sin(\pi/3) = \sqrt{3}/2\end{align*}, and \begin{align*}f'(x) = \cos x,\end{align*} \begin{align*}f'(\pi/3) = \cos(\pi/3) = 1/2,\end{align*} we have

\begin{align*}f(x) & \approx \frac{\sqrt{3}} {2} + \frac{1} {2} \left (x - \frac{\pi} {3} \right)\\ & \approx \frac{\sqrt{3}} {2} + \frac{x} {2} - \frac{\pi} {6}\\ & \approx \frac{x} {2} + 0.343.\end{align*}

Figure 10

## Newton’s Method

When faced with a mathematical problem that cannot be solved with simple algebraic means, such as finding the roots of the polynomial \begin{align*}x^3 -2x + 3 = 0,\end{align*} calculus sometimes provides a way of finding the approximate solutions.

Let's say we are interested in computing \begin{align*} \sqrt{5}\end{align*} without using a calculator or a table. To do so, think about this problem in a different way. Assume that we are interested in solving the quadratic equation

\begin{align*}f(x) = x^2 - 5 = 0\end{align*}

which leads to the roots \begin{align*} x = \pm \sqrt{5}\end{align*}.

The idea here is to find the linearization of the above function, which is a straight-line equation, and then solve the linear equation for \begin{align*}x\end{align*}.

Since

\begin{align*} \sqrt{4} < \sqrt{5} < \sqrt{9}\end{align*}

or

\begin{align*} 2 < \sqrt{5} < 3,\end{align*}

We choose the linear approximation of \begin{align*}f(x)\end{align*} to be near \begin{align*}x_0 = 2\end{align*}. Since \begin{align*}f(x) = x^2 - 5,\end{align*} \begin{align*}f'(x) = 2x\end{align*} and thus \begin{align*}f(2) = -1\end{align*} and \begin{align*}f'(2) = 4.\end{align*} Using the linear approximation formula,

\begin{align*}f(x) & \approx f(x_0) + f'(x_0)(x - x_0)\\ & \approx -1 + (4)(x - 2)\\ & \approx -1 + 4x - 8\\ & \approx 4x - 9.\end{align*}

Notice that this equation is much easier to solve than \begin{align*}f(x) = x^2 - 5.\end{align*} Setting \begin{align*}f(x) = 0\end{align*} and solving for \begin{align*}x\end{align*}, we obtain,

\begin{align*}4x - 9 & = 0\\ x & = \frac{9} {4}\\ & = 2.25.\end{align*}

If you use a calculator, you will get \begin{align*}x = 2.236 \ldots\end{align*} As you can see, this is a fairly good approximation. To be sure, calculate the percent difference \begin{align*}[\%\;\mathrm{diff}]\end{align*} between the actual value and the approximate value,

\begin{align*}\%\ \text{diff} = \frac{2|A - X|} {|A + X|} 100\%,\end{align*}

where \begin{align*}A\end{align*} is the accepted value and \begin{align*}X\end{align*} is the calculated value.

\begin{align*}\%\ \text{diff} &= \frac{2|2.236 - 2.25|} {|2.236 + 2.25|} 100\%\\ &= 0.62\%,\end{align*}

which is less than \begin{align*}1\% \end{align*}.

We can actually make our approximation even better by repeating what we have just done not for \begin{align*}x = 2\end{align*}, but for \begin{align*}x_1 = 2.25 = \frac{9} {4}\end{align*}, a number that is even closer to the actual value of \begin{align*} \sqrt{5}\end{align*}. Using the linear approximation again,

\begin{align*}f(x)& \approx f(x_1) + f'(x_1)(x - x_1)\\ & \approx \frac{1} {16} + \frac{9} {2} \left (x - \frac{9} {4} \right)\\ & \approx \frac{9} {2}x - \frac{161} {16}.\end{align*}

Solving for \begin{align*}x\end{align*} by setting \begin{align*}f(x) = 0\end{align*}, we obtain

\begin{align*}x = x_2 = 2.236111,\end{align*}

which is even a better approximation than \begin{align*}x_1 = 9/4\end{align*}. We could continue this process generating a better approximation to \begin{align*} \sqrt{5}\end{align*}. This is the basic idea of Newton’s Method.

Here is a summary of Newton’s method.

Newton’s Method

1. Guess the first approximation to a solution of the equation \begin{align*}f(x) = 0\end{align*}. A graph would be very helpful in finding the first approximation (see figure below).
2. Use the first approximation to find the second, the second to find the third and so on by using the recursion relation

\begin{align*} x_{n+1} = x_n - \frac{f(x_n)} {f'(x_n)}.\end{align*}

Example 3:

Use Newton’s method to find the roots of the polynomial \begin{align*}f(x) = x^3 + x - 1.\end{align*}

Solution:

\begin{align*}f(x) &= x^3 + x - 1\\ f'(x) &= 3x^2 + 1.\end{align*}

Using the recursion relation,

\begin{align*}x_{n+1}& = x_n - \frac{f(x_n)} {f'(x_n)}\\ & = x_n - \frac{x^3_n + x_n - 1} {3x^2_n + 1}.\end{align*}

To help us find the first approximation, we make a graph of \begin{align*}f(x)\end{align*}. As Figure 11 suggests, set \begin{align*}x_1 = 0.6\end{align*}. Then using the recursion relation, we can generate \begin{align*}x_2\end{align*}, \begin{align*}x_3, \ldots\end{align*}.

Figure 11

\begin{align*}x_{n+1} & = x_n - \frac{x^3_n + x_n - 1} {3x^2_n + 1}\\ x_2 & = 0.6 - \frac{(0.6)^3 + (0.6) - 1} {3(0.6)^2 + 1}\\ & = 0.6884615.\end{align*}

Using the recursion relation again to find \begin{align*}x_3\end{align*}, we get

\begin{align*}x_3 = 0.6836403.\end{align*}

We conclude that the solution to the equation \begin{align*}x^3 + x - 1 = 0\end{align*} is about \begin{align*}0.6836403\end{align*}.

## Multimedia Links

For a video presentation of Newton's method (10.0), see Math Video Tutorials by James Sousa, Newton's method (9:48).

## Review Questions

1. Find the linearization of \begin{align*} f(x) = \frac{x^2 + 1} {x}\end{align*} at \begin{align*}a = 1\end{align*}.
2. Find the linearization of \begin{align*}f(x) = \tan x\end{align*} at \begin{align*}a = \pi\end{align*}.
3. Use the linearization method to show that when \begin{align*}x \ll 1\end{align*} (much less than \begin{align*}1\end{align*}), then \begin{align*}(1 + x)^n \approx 1 + nx\end{align*}. Hint: Let \begin{align*}x = 0.\end{align*}
4. Use the result of problem #3, \begin{align*}(1 + x)^n \approx 1 + nx\end{align*}, to find the approximation for the following:
1. \begin{align*}f(x) = (1 - x)^4\end{align*}
2. \begin{align*} f(x) = \sqrt{1 - x}\end{align*}
3. \begin{align*} f(x) = \frac{5} {\sqrt{1 + x}}\end{align*}
4. Without using a calculator, approximate \begin{align*}(1.003)^{99}\end{align*}.
5. Use Newton’s Method to find the roots of \begin{align*}x^3 + 3 = 0\end{align*}.
6. Use Newton’s Method to find the roots of \begin{align*} -x + 3\sqrt{-1 + x} = 0\end{align*}.

## Texas Instruments Resources

In the CK-12 Texas Instruments Calculus FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/book/CK-12-Texas-Instruments-Calculus-Student-Edition/section/3.0/.

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