# 2.7: Linearization and Newton’s Method

**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 \begin{align*}f\end{align*} is a differentiable function at \begin{align*}x_0\end{align*}, then the tangent line, \begin{align*}y = mx + b\end{align*}, to the curve \begin{align*}y = f(x)\end{align*} at \begin{align*}x_0\end{align*} is a good approximation to the curve \begin{align*}y = f(x)\end{align*} for values of \begin{align*}x\end{align*} near \begin{align*}x_0\end{align*} (Figure 8a). If you “zoom in” on the two graphs, \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**

- 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).
- 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

- Find the linearization of \begin{align*} f(x) = \frac{x^2 + 1} {x}\end{align*} at \begin{align*}a = 1\end{align*}.
- Find the linearization of \begin{align*}f(x) = \tan x\end{align*} at \begin{align*}a = \pi\end{align*}.
- 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*}
- Use the result of problem #3, \begin{align*}(1 + x)^n \approx 1 + nx\end{align*}, to find the approximation for the following:
- \begin{align*}f(x) = (1 - x)^4\end{align*}
- \begin{align*} f(x) = \sqrt{1 - x}\end{align*}
- \begin{align*} f(x) = \frac{5} {\sqrt{1 + x}}\end{align*}
- Without using a calculator, approximate \begin{align*}(1.003)^{99}\end{align*}.

- Use Newton’s Method to find the roots of \begin{align*}x^3 + 3 = 0\end{align*}.
- 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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Feb 23, 2012## Last Modified:

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