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# 3.8: Approximation Errors

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Extend the Mean Value Theorem to make linear approximations.
• Analyze errors in linear approximations.
• Extend the Mean Value Theorem to make quadratic approximations.
• Analyze errors in quadratic approximations.

## Introduction

In this lesson we will use the Mean Value Theorem to make approximations of functions. We will apply the Theorem directly to make linear approximations and then extend the Theorem to make quadratic approximations of functions.

Let’s consider the tangent line to the graph of a function f\begin{align*}f\end{align*} at the point (a,f(a)).\begin{align*}(a, f(a)).\end{align*} The equation of this line is y=f(a)+f(a)(xa).\begin{align*}y = f(a) + f'(a)(x - a).\end{align*} We observe from the graph that as we consider x\begin{align*}x\end{align*} near a,\begin{align*}a,\end{align*} the value of f(x)\begin{align*}f(x)\end{align*} is very close to f(a).\begin{align*}f(a).\end{align*}

In other words, for x\begin{align*}x\end{align*} values close to a,\begin{align*}a,\end{align*} the tangent line to the graph of a function f\begin{align*}f\end{align*} at the point (a,f(a))\begin{align*}(a, f(a))\end{align*} provides an approximation of f(x)\begin{align*}f(x)\end{align*} or f(x)f(a)+f(a)(xa).\begin{align*}f(x) \approx f(a) + f'(a)(x-a).\end{align*} We call this the linear or tangent line approximation of f\begin{align*}f\end{align*} at a\begin{align*}a\end{align*} and indicate it by the formula L(x)=f(a)+f(a)(xa).\begin{align*}L(x) = f(a) + f'(a)(x-a).\end{align*}

The linear approximation can be used to approximate functional values that deviate slightly from known values. The following example illustrates this process.

Example 1:

Use the linear approximation of the function f(x)=x2\begin{align*}f(x)=\sqrt{x-2}\end{align*} at a=6\begin{align*}a = 6\end{align*} to approximate 3.95\begin{align*}\sqrt{3.95}\end{align*}.

Solution:

We know that f(6)=62=2\begin{align*} f(6)=\sqrt{6-2}=2\end{align*}. So we will find the linear approximation of the function and substitute x\begin{align*}x\end{align*} values close to 6\begin{align*}6\end{align*}

L(x)=f(6)+f(6)(x6).

We note that f(x)=12x2,f(6)=14.\begin{align*} f'(x)=\frac {1}{2 \sqrt{x-2}},f'(6)=\frac {1}{4}.\end{align*}

We also know that f(6)=2.\begin{align*}f (6) = 2.\end{align*}

By substitution, we have

f(x)f(6)+f(6)(x6)\begin{align*}f(x)\approx f(6) + f'(6)(x - 6)\end{align*} for x\begin{align*}x\end{align*} near 6\begin{align*}6\end{align*}

Hence L(x)=2+14(x6)=12+14x.\begin{align*} L(x)=2+\frac{1}{4} (x-6)=\frac{1}{2}+\frac{1}{4}x.\end{align*}

We observe that to approximate 3.95\begin{align*}\sqrt{3.95}\end{align*} we need to evaluate the linear approximation at 5.95\begin{align*}5.95\end{align*}, and we have

L(5.95)=12+14(5.95)=1.9875\begin{align*} L(5.95)=\frac{1}{2}+\frac{1}{4} (5.95)=1.9875\end{align*}. If we were to compare this approximation to the actual value, 3.951.9874\begin{align*}\sqrt {3.95}\approx 1.9874\end{align*}, we see that it is a very good approximation.

If we observe a table of x\begin{align*}x\end{align*} values close to 6,\begin{align*}6,\end{align*} we see how the approximations compare to the actual value.

f(x)=x23.953.9944.14.05x5.955.9966.016.05L(x)=12+14x1.98751.997522.00252.0125Actual1.98741.997422.00242.0124

Setting Error Estimates

We would like to have confidence in the approximations we make. We therefore can choose the x\begin{align*}x\end{align*} values close to a to ensure that the errors are within acceptable boundaries. For the previous example, we saw that the values of L(x)\begin{align*}L(x)\end{align*} close to a=6,\begin{align*}a = 6,\end{align*} gave very good approximations, all within 0.0001\begin{align*}0.0001\end{align*} of the actual value.

Example 2:

Let’s suppose that for the previous example, we did not require such precision. Rather, suppose we wanted to find the range of x\begin{align*}x\end{align*} values close to 6\begin{align*}6\end{align*} that we could choose to ensure that our approximations lie within 0.01\begin{align*}0.01\end{align*} of the actual value.

Solution:

The easiest way for us to find the proper range of x\begin{align*}x\end{align*} values is to use the graphing calculator. We first note that our precision requirement can be stated as |x2(12+x4)|<0.01.\begin{align*}|\sqrt{x-2}-\left (\frac{1}{2}+\frac{x}{4}\right )|<0.01.\end{align*}

If we enter the functions \begin{align*}f(x)=\sqrt{x-2}\end{align*} and \begin{align*}L(x)=\frac{1}{2}+\frac{1}{4} x\end{align*} into the \begin{align*}Y =\end{align*} menu as \begin{align*}Y_1\end{align*} and \begin{align*}Y_2\end{align*}, respectively, we will be able to view the function values of the functions using the [TABLE] feature of the calculator. In order to view the differences between the actual and approximate values, we can enter into the \begin{align*}Y =\end{align*} menu the difference function \begin{align*}Y_3 = Y_1 - Y_2\end{align*} as follows:

1. Go to the \begin{align*}Y =\end{align*} menu and place cursor on the \begin{align*} Y_3\end{align*} line.
2. Press the following sequence of key strokes: [VARS] [FUNCTION] \begin{align*}[Y_1]\end{align*}. This will copy the function \begin{align*}Y_1\end{align*} onto the \begin{align*}Y_3\end{align*} line of the \begin{align*}Y =\end{align*} menu.
3. Press [-] to enter the subtraction operation onto the \begin{align*}Y_3\end{align*} line of the \begin{align*}Y =\end{align*} menu.
4. Repeat steps 1 - 2 and choose \begin{align*}Y_2\end{align*} to copy \begin{align*}Y_2\end{align*} onto the \begin{align*}Y_3\end{align*} line of the \begin{align*}Y =\end{align*} menu.

Your screen should now appear as follows:

Now let’s setup the [TABLE] function so that we find the required accuracy.

1. Press 2ND followed by [TBLSET] to access the Table Setup screen.
2. Set the [TBLStart] value to \begin{align*}5\end{align*} and \begin{align*}\triangle \;\mathrm{Tbl}\end{align*} to \begin{align*}0.1\end{align*}.

Your screen should now appear as follows:

Now we are ready to find the required accuracy.

Access the [TABLE] function, scroll through the table, and find those \begin{align*}x\end{align*} values that ensure \begin{align*}|Y_3| \le 0.01\end{align*}.

\begin{align*}5.3 \le x \le 6.8\end{align*}

Non-Linear Approximations

It turns out that the linear approximations we have discussed are not the only approximations that we can derive using derivatives. We can use non-linear functions to make approximations. These are called Taylor Polynomials and are defined as

We call this the Taylor Polynomial of fcentered at a.

For our discussion, we will focus on the quadratic case. The Taylor Polynomial corresponding to \begin{align*}n = 2\end{align*} is given by

Note that this is just our linear approximation with an added term. Hence we can view it as an approximation of \begin{align*}f\end{align*} for \begin{align*}x\end{align*} values close to \begin{align*}a.\end{align*}

Example 3:

Find the quadratic approximation of the function \begin{align*}f(x) = \sqrt{x - 2}\end{align*} at \begin{align*}a = 6\end{align*} and compare them to the linear approximations from the first example.

Solution:

Recall that \begin{align*} L(x) = \frac{1} {2} + \frac{1} {4}x.\end{align*}

Hence \begin{align*} T_2(x) = L(x) + \frac{1} {2} f'' (6) (x - 6)^2\end{align*}.

\begin{align*} f''(x) = -\frac{1} {4\sqrt{(x - 2)^3}}\end{align*} ; so \begin{align*} f''(6) = -\frac{1} {4\sqrt{(6 - 2)^3}} = - \frac{1} {32}.\end{align*}

Hence \begin{align*}T_2(x) = L(x) - \frac{1} {64} (x - 6)^2 = \frac{1} {2} + \frac{1} {4}x - \frac{1} {64} (x - 6)^2 = - \frac{1} {64} x^2 + \frac{7} {16} x - \frac {1} {16}.\end{align*}

So \begin{align*}T_2(x) = - \frac{1} {64} x^2 + \frac{7} {16} x - \frac{1} {16}\end{align*}. If we update our table from the first example we can see how the quadratic approximation compares with the linear approximation.

As you can see from the graph below, \begin{align*}T(x)\end{align*} is an excellent approximation of \begin{align*}f(x)\end{align*} near \begin{align*}x = 6.\end{align*}

We get a slightly better approximation for the quadratic than for the linear. If we reflect on this a bit, the finding makes sense since the shape and properties of quadratic functions more closely approximate the shape of radical functions.

Finally, as in the first example, we wish to determine the range of \begin{align*}x\end{align*} values that will ensure that our approximations are within \begin{align*}0.01\end{align*} of the actual value. Using the [TABLE] feature of the calculator, we find that if \begin{align*}4.444 \le \times \le 7.87,\end{align*} then \begin{align*} |\sqrt{x -2} - T(x)| < 0.01\end{align*}.

## Lesson Summary

1. We extended the Mean Value Theorem to make linear approximations.
2. We analyzed errors in linear approximations.
3. We extended the Mean Value Theorem to make quadratic approximations.
4. We analyzed errors in quadratic approximations.

## Review Questions

In problems #1–4, find the linearization \begin{align*}L(x)\end{align*} of the function at \begin{align*}x = a.\end{align*}

1. \begin{align*}f(x) = 2x^4 - 6x^3\end{align*} near \begin{align*}a = -2\end{align*}
2. \begin{align*} f(x) = x^\frac{2} {3}\end{align*} near \begin{align*}a = 27\end{align*}
3. Find the linearization of the function \begin{align*} f(x) = \sqrt{5 -x}\end{align*} near a = 1 and use it to approximate \begin{align*}\sqrt{4.01}\end{align*}.
4. Based on using linear approximations, is the following approximation reasonable?
5. Use a linear approximation to approximate the following:
6. Verify the the following linear approximation at \begin{align*}a = 1.\end{align*} Determine the values of \begin{align*}x\end{align*} for which the linear approximation is accurate to \begin{align*}0.01.\end{align*}
7. Find the quadratic approximation for the function in #3, \begin{align*} f(x) = \sqrt{5 - x}\end{align*} near \begin{align*}a = 1.\end{align*}
8. Determine the values of \begin{align*}x\end{align*} for which the quadratic approximation found in #7 is accurate to \begin{align*}0.01.\end{align*}
9. Determine the quadratic approximation for \begin{align*}f(x) = 2x^4 - 6x^3\end{align*} near \begin{align*}a = -2.\end{align*} Do you expect that the quadratic approximation is better or worse than the linear approximation? Explain your answer.

1. \begin{align*}f(-2) = 80,\end{align*} \begin{align*}f'(-2) = -136\end{align*}; \begin{align*}L(x) = -192 -136x\end{align*}
2. \begin{align*}f(27) = 9,\end{align*} \begin{align*}f'(27) = 2/9\end{align*}; \begin{align*}L(x) = 3 + \frac{2}{9} x\end{align*}
3. \begin{align*}f(1) = 2,\end{align*} \begin{align*}f'(1) = -\frac{1} {4}\end{align*}; \begin{align*}L(x) = \frac{9}{4} - \frac{1} {4} x\end{align*}; \begin{align*}\sqrt{4.01} \approx 2.0025\end{align*}
4. Yes; using linear approximation on \begin{align*}f(x) = x^4\end{align*} near \begin{align*}a = 1\end{align*} we find that \begin{align*}L(x) = 4x -3\end{align*}; \begin{align*}L(1.001) = 1.004.\end{align*}
5. Using linear approximation on \begin{align*}f(x) = x^\frac{3} {4}\end{align*} near \begin{align*}a = 16\end{align*} we find \begin{align*}L(x) = 8+ \frac{3} {8} (x - 16) = 2 + \frac{3} {8}x\end{align*}; \begin{align*}L(16.08) = 8.03.\end{align*}
6. \begin{align*}0.68 < \times < 1.27\end{align*}
7. \begin{align*} T(x) = - \frac{1} {64} x^2 - \frac{7} {32} x + \frac{143} {64}\end{align*}
8. \begin{align*}0.87 \le \times \le 2.56\end{align*}
9. \begin{align*}T(x) = 84x^2 + 200x + 144\end{align*}; we would expect it to be a better approximation since the graph of \begin{align*}f(x) = 2x^4 - 6x^3\end{align*} is similar to the graph of a quadratic function.

## Texas Instruments Resources

In the CK-12 Texas Instruments Calculus FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9728.

Feb 23, 2012

Oct 30, 2015