AC Extending local linearization

Section 8.1 Extending local linearization

Motivating Questions

How well does the tangent line at $a = 0$ approximate the function $f (x) = e^{x}$ near $a = 0 ?$
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Can we find higher degree polynomials that approximate $f (x) = e^{x}$ near $a = 0$ more effectively than the approximation generated by the tangent line?
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How does the degree of the polynomial impact the accuracy of the approximation of $f (x) = e^{x} ?$
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Subsection 8.1.1 Introduction

Early in our study of calculus in Section 1.8, we learned that if a function

f

has a derivative at a fixed value

x = a,

when we zoom in on its graph near

(a, f (a)),

the function looks linear. Indeed, such a function is differentiable, and we know that near a fixed input value

a,

f (x) \approx L (x) = f (a) + f^{'} (a) (x - a),

where

L

is the tangent line approximation to

f

a .

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In this section, we use the function

f (x) = e^{x}

as a case study to investigate how we can use other basic functions to better approximate the value of

f (x)

near

a = 0 .

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Preview Activity 8.1.1.

Consider the function

f (x) = e^{x}

near

a = 0 .

We know that

f^{'} (x) = e^{x},

f^{'} (0) = 1;

along with the fact that

f (0) = 1,

it follows that the tangent line approximation is

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L (x) = f (0) + f^{'} (0) (x - 0) = 1 + 1 (x - 0) = 1 + x .

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Here is the function

f (x) = e^{x}

and its tangent line

L (x) = 1 + x

near the point

(0, f (0)) .

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(a)

Build a spreadsheet that computes the difference between

f (x)

and

L (x)

for

x

-values between

- 1

and

1,

spaced

0.1

units apart. Note: we will revisit this spreadsheet for additional computations in Activity 8.1.4, so it would be ideal if you save your work electronically in place you can find it later.

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Your spreadsheet should start like the following table:

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Row	$Δ x$	$x$	$f (x) = e^{x}$	$L (x) = 1 + x$
0	$0.1$	$- 1.0$	$0.36787 \dots$	$0.0$
1	$0.1$	$- 0.9$	$0.40656 \dots$	$0.1$
2	$0.1$	$- 0.8$	$0.44932 \dots$	$0.2$
⋮	⋮	⋮	⋮	⋮

Then Row 14 is:

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Row	$Δ x$	$x$	$f (x) = e^{x}$	$L (x) = 1 + x$
14

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(b)

Next, add a column to your spreadsheet that computes

| f (x) - L (x) | .

What is the new entry in Row 14?

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How do the differences between

f (x)

and

L (x)

change as you move across the interval from

- 1

1 ?

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No changes.
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Differences get smaller and smaller.
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Differences get larger and larger.
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Differences get larger the farther away $x$ is from $0 .$
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(c)

For about what values of

x

is it true that

| f (x) - L (x) | < 0.1 ?

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(d)

Notice that the curvature in

f (x) = e^{x}

is what makes the linear approximation

L (x)

lose accuracy. What kind of simple function might do a better job approximating

e^{x}

than a linear one?

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quadratic
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sine
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cosine
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logarithm
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Subsection 8.1.2 Finding a quadratic approximation

In Preview Activity 8.1.1, we found that the error in the tangent line approximation of

f (x) = e^{x}

a = 0

grows significantly as we consider

x

-values further and further from

0 .

This is due to the fact that the tangent line is straight while the function

f (x) = e^{x}

has some curvature. To hopefully improve the approximation, we are going to try to find a quadratic function whose curvature matches that of

f (x) = e^{x}

at the point of tangency.

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While we have usually used the notation “

L (x)

” for the tangent line, in what follows we will instead write “

T_{1} (x)

”, and think of this as “the degree

1

approximation”. In a similar way, we will write “

T_{2} (x)

” for the quadratic (degree

2

) approximation.

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Recall that for any function

f

that has a derivative at

a = 0,

its tangent line approximation at

a = 0

T_{1} (x) = f (0) + f^{'} (0) (x - 0) .

Moreover, the functions

T_{1}

and

f

have two exact values in common. First, their function values agree at the point of tangency:

T_{1} (0) = f (0) .

And second, since

T_{1} (x)

is a linear function whose slope is

f^{'} (0),

it is also true that their derivative values agree at the point of tangency:

T_{1}^{'} (0) = f^{'} (0) .

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To generate a quadratic function that approximates

f

near

a = 0,

we choose to have this quadratic function not only share the same function value and derivative value as

f

a = 0,

but also the same second derivative value

Here we are implicitly assuming that the function

f (x)

has a second derivative at

a = 0,

which is a property that holds for

f (x) = e^{x} .

a = 0

in order to match the concavity or curvature of

f .

In other words, we are adding a term to the linear approximation that gives the same amount of curvature as the function

f .

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We can state these requirements more formally as follows.

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The quadratic approximation of $f (x) = e^{x}$ .

To extend the linear approximation of

f (x) = e^{x}

to a quadratic approximation, we seek a function

T_{2} (x)

of the form

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2}

that satisfies

$T_{2} (0) = f (0),$ so $T_{2}$ and $f$ share the same height at $a = 0;$
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$T_{2}^{'} (0) = f^{'} (0),$ so $T_{2}$ and $f$ share the same slope at $a = 0;$
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$T_{2}^{″} (0) = f^{″} (0),$ so $T_{2}$ and $f$ share the same concavity at $a = 0 .$
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In Activity 8.1.2, we explore how these three requirements determine

b_{0},

b_{1},

and

b_{2}

T_{2} (x)

for the function

f (x) = e^{x} .

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Activity 8.1.2.

Let

f (x) = e^{x}

and

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} .

We seek numerical values for the constants

b_{0},

b_{1},

and

b_{2}

so that

f (0) = T_{2} (0),

f^{'} (0) = T_{2}^{'} (0),

and

f^{″} (0) = T_{2}^{″} (0) .

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(a)

Note that since

b_{0},

b_{1},

and

b_{2}

are constants, if we take the derivative of the quadratic function

T_{2}

using the sum and constant multiple rules, it follows that

T_{2}^{'} (x) = b_{1} + 2 b_{2} x .

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What is

T_{2}^{″} (x) ?

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(b)

Recall that

f (x) = e^{x} .

Determine

f^{'} (x)

and

f^{″} (x) .

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(c)

Enter the formulas you’ve determined for

f^{'} (x),

f^{″} (x),

and

T_{2}^{″} (x)

to fill in the blanks below.

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\begin{aligned} f (x) & = e^{x} & T_{2} (x) & = b_{0} + b_{1} x + b_{2} x^{2} \\ f^{'} (x) & = & T_{2}^{'} (x) & = b_{1} + 2 b_{2} x \\ f^{″} (x) & = & T_{2}^{″} (x) & = \end{aligned}

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(d)

Next, observe that since

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2},

it follows that

T_{2} (0) = b_{0} .

Reason similarly to determine the values of

T_{2}^{'} (0)

and

T_{2}^{″} (0),

as well as those of

f (0),

f^{'} (0),

and

f^{″} (0)

and enter these values appropriately in the blanks below.

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\begin{aligned} f (0) & = & T_{2} (0) & = b_{0} \\ f^{'} (0) & = & T_{2}^{'} (0) & = \\ f^{″} (0) & = & T_{2}^{″} (0) & = \end{aligned}

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(e)

Now, recall that we want the function values, first derivative values, and second derivative values of

f

and

T_{2}

to match at

a = 0 .

What does

T_{2} (0) = f (0)

tell us about the value of

b_{0},

and what is its value? What does

T_{2}^{'} (0) = f^{'} (0)

imply the value of

b_{1}

is? How can we reason similarly to find

b_{2} ?

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(f)

Having now determined the numerical values of

b_{0},

b_{1},

and

b_{2},

use appropriate computing technology to plot the function

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2}

along with

f (x) = e^{x}

and

T_{1} (x) = 1 + x

in the same window as that shown in Figure 8.1.3.

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Figure 8.1.3. The function $f (x) = e^{x}$ and its tangent line $T_{1} (x) = 1 + x$ near the point $(0, f (0)) .$
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What do you notice? For about which values of

x

| f (x) - T_{2} (x) | < 0.1 ?

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Subsection 8.1.3 Over and over again

A remarkable feature of mathematics is that when a process effectively generates an approximation, doing that same process again (perhaps with some slight modifications) often improves the approximation. In Activity 8.1.2, we found a quadratic approximation of

f (x) = e^{x}

near the point

(0, f (0))

that results in an improvement over the linear approximation of

f .

It is reasonable to hope that a degree 3 polynomial approximation of

f (x) = e^{x}

will be even better.

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To investigate, we seek a degree

3

polynomial

T_{3} (x)

of the form

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3}

that satisfies

the same conditions we imposed on $T_{2} (x) :$

$T_{3} (0) = f (0), T_{3}^{'} (0) = f^{'} (0), and T_{3}^{″} (0) = f^{″} (0)$

so that $T_{3}$ and $f$ share the same function value, first derivative value, and second derivative value at $a = 0,$

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plus the additional condition that

$T_{3}^{‴} (0) = f^{‴} (0),$

so $T_{3}$ and $f$ share the same third derivative value
²
Here we are assuming that the original function $f$ has a third derivative at $a = 0,$ which is valid since $f^{‴} (x) = e^{x} .$
at $a = 0 .$

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We are using the unknown constants

c_{0},

c_{1},

c_{2},

and

c_{3}

for

T_{3} (x)

instead of the constants

b_{0},

b_{1},

and

b_{2}

that we considered for

T_{2} (x)

since we don’t yet know whether the first three values of

c_{i}

will be the same as those of

b_{i}

or not.

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Like in our work with

T_{2},

we observe that since

T_{3}

is a polynomial, its derivatives are straightforward to compute. For instance,

T_{3}^{'} (x) = c_{1} + 2 c_{2} x + 3 c_{3} x^{2} .

We continue our investigation of this new approximation of

f (x) = e^{x}

in Activity 8.1.3, where we work to determine the values of

c_{0},

c_{1},

c_{2},

and

c_{3}

plus explore how well

T_{3} (x)

approximates

f (x)

near

a = 0 .

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Activity 8.1.3.

Let

f (x) = e^{x}

and

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} .

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(a)

By computing the third derivative of

f (x)

and the second and third derivatives of

T_{3} (x)

and evaluating the relevant functions at

x = 0,

fill in the blanks below.

\begin{aligned} f (x) & = e^{x} & T_{3} (x) & = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} \\ f^{'} (x) & = e^{x} & T_{3}^{'} (x) & = c_{1} + 2 c_{2} x + 3 c_{3} x^{2} \\ f^{″} (x) & = e^{x} & T_{3}^{″} (x) & = \\ f^{‴} (x) & = & T_{3}^{‴} (x) & = \\ f (0) & = 1 & T_{3} (0) & = \\ f^{'} (0) & = 1 & T_{3}^{'} (0) & = \\ f^{″} (0) & = 1 & T_{3}^{″} (0) & = \\ f^{‴} (0) & = & T_{3}^{‴} (0) & = \end{aligned}

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(b)

Next, recall that we want

f

and

T_{3}

to share the same function and derivative values at

a = 0

up to and including the third derivative. For instance, one of the four needed equations is

T_{3}^{'} (0) = f^{'} (0) .

Use the four equations your work in the preceding question to determine the values of

c_{0},

c_{1},

c_{2},

and

c_{3} .

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(c)

Having now determined the numerical values of

c_{0},

c_{1},

c_{2},

and

c_{3},

use appropriate computational technology to plot the function

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3}

along with

f (x) = e^{x},

T_{1} (x) = 1 + x,

and

T_{2} (x) = 1 + x + \frac{1}{2} x^{2}

in the same window as shown in Figure 8.1.7.

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Figure 8.1.7. The function $f (x) = e^{x},$ its tangent line $T_{1} (x) = 1 + x,$ and the quadratic approximation $T_{2} (x) = 1 + x + \frac{1}{2} x^{2}$ near the point $(0, f (0)) .$
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What do you notice? For approximately which values of

x

| f (x) - T_{3} (x) | < 0.1 ?

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(d)

What if we wanted a degree-

4

polynomial approximation to

f (x) = e^{x}

near

a = 0 ?

Based on the patterns you’ve observed in

T_{1},

T_{2},

and

T_{3},

conjecture values for the constants

d_{0}, \dots, d_{4}

for a function

T_{4}

of the form

T_{4} (x) = d_{0} + d_{1} x + d_{2} x^{2} + d_{3} x^{3} + d_{4} x^{4}

that satisfies

T_{4} (0) = f (0),

T_{4}^{'} (0) = f^{'} (0),

\dots,

T_{4}^{(4)} (0) = f^{(4)} (0) .

Add this function

T_{4}

to your plot in part (c) that includes

f (x)

and the lower-degree polynomial approximations. What do you notice?

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Subsection 8.1.4 As the degree of the approximation increases

Our work so far with the case study function

f (x) = e^{x}

suggests that as we find degree

n

polynomial approximations,

T_{n},

that satisfy

T_{n} (0) = f (0), T_{n}^{'} (0) = f^{'} (0), T_{n}^{″} (0) = f_{n}^{″} (0), \dots, T_{n}^{(n)} (0) = f^{(n)} (0),

increasing the value of

n

improves the accuracy of the approximation.

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In the next activity, we introduce the idea of the error of a polynomial approximation and investigate explicitly how the error varies for approximations of

f (x) = e^{x}

as we vary

n

and vary

x .

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Activity 8.1.4.

We continue to work with

f (x) = e^{x}

and the four approximations of degree

1,

2,

3,

and

4

given by

T_{1} (x),

T_{2} (x),

T_{3} (x),

and

T_{4} (x) .

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(a)

In Preview Activity 8.1.1, we built a spreadsheet that computed the differences between

f (x)

and

T_{1} (x)

for

x

-values between

- 1

and

1,

spaced

0.1

units apart. Your spreadsheet started like the one shown in the table in Preview Activity 8.1.1.

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Next, we build an updated version of this spreadsheet that computes similar differences between

f

and the three higher degree approximations we have found. In particular, we now want to have columns for

Δ x,

x,

f (x),

T_{1} (x),

T_{2} (x),

T_{3} (x),

and

T_{4} (x),

plus the absolute differences

| f (x) - T_{1} (x) |,

| f (x) - T_{2} (x) |,

| f (x) - T_{3} (x) |,

and

| f (x) - T_{4} (x) | .

Hint: when building your entries, note that you can think of

T_{2} (x)

T_{2} (x) = T_{1} (x) + \frac{1}{2} x^{2},

and similarly view

T_{3} (x)

as “

T_{2} (x)

plus one more term”.

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Include at least

5

digits of accuracy beyond the decimal. The first seven columns of your spreadsheet might start like this:

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Table 8.1.9. Comparing

f (x) = e^{x}

and its degree

1,

2,

3,

4

and approximations near

a = 0 .

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$Δ x$	$x$	$f (x)$	$T_{1} (x)$	$T_{2} (x)$	$T_{3} (x)$	$T_{4} (x)$
$0.1$	$- 1.0$	$0.36787$	$0.00000$	$0.50000$	$0.33333$	$0.37500$
$0.1$	$- 0.9$	$0.40657$	$0.10000$	$0.50500$	$0.38350$	$0.41083$

The next four columns of your spreadsheet should begin as follows:

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Table 8.1.10. The absolute error between

f (x) = e^{x}

and its degree

1,

2,

3,

and

4

approximations at

x = - 1

and

x = - 0.9 .

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$\| f (x) - T_{1} (x) \|$	$\| f (x) - T_{2} (x) \|$	$\| f (x) - T_{3} (x) \|$	$\| f (x) - T_{4} (x) \|$
$0.36787$	$0.13212$	$0.03454$	$0.00712$
$0.30657$	$0.09843$	$0.02307$	$0.00426$

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(b)

We call the value of

| f (x) - T_{2} (x) |

the absolute error of the quadratic approximation of

f

at the value

x

. What is the absolute error of the quadratic approximation at

x = - 1 ?

x = 1 ?

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(c)

What is the absolute error of the cubic (degree

3

) approximation,

T_{3} (x),

x = - 1 ?

x = 1 ?

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(d)

What is the absolute error of the quartic (degree

4

) approximation,

T_{4} (x),

x = - 1 ?

x = 1 ?

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(e)

Study your spreadsheet for trends that you notice as the value of

x

changes or the degree

n

of the approximation changes. What are your observations?

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(f)

Investigate the errors in the various approximations for a wider interval of

x

-values. For example, you might consider starting at

x = - 2

with

Δ x = 0.2 .

What do you notice?

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One important application of our work so far is that these polynomial approximations provide a way to approximate values of the function

f (x) = e^{x} .

For example, since we’ve shown that

e^{x} \approx 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4},

it follows that

e^{\frac{1}{2}} \approx 1 + \frac{1}{2} + \frac{1}{2} {(\frac{1}{2})}^{2} + \frac{1}{6} {(\frac{1}{2})}^{3} + \frac{1}{24} {(\frac{1}{2})}^{4} = \frac{211}{128} = 1.6484375 .

In fact, this approach through polynomial approximation is one way that computers determine the value of

e^{\frac{1}{2}},

which is approximately

1.64872127,

to whatever accuracy is needed: by using even better polynomial approximations than the degree-

4

one that we found, computers are able to generate the approximate value

1.64872127

simply by the basic computations of addition and multiplication with enough terms.

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Throughout this section, we have focused on

f (x) = e^{x} .

One of the characteristics that makes

f (x) = e^{x}

special is the fact that its derivative is itself; indeed, the

n^{th}

derivative of

f

f^{(n)} (x) = e^{x}

for every natural number

n,

which in turn implies that

f^{(n)} (0) = 1

for every value of

n .

This will ultimately help to find patterns in the coefficients of the degree

n

polynomial approximation,

T_{n} (x),

and be able to easily write down a formula for any value of

n .

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It is natural to think that we can find similar approximations of other functions, especially ones such as

\sin (x)

and

\cos (x)

that also exhibit repeating patterns in their derivatives. In Section 8.2, we will develop a general approach to finding the coefficient of

x^{n}

in the degree

n

approximation of any function with

n

derivatives and learn how to find a general expression for the degree

n

approximation.

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Subsection 8.1.5 Summary

For the function $f (x) = e^{x},$ which bends considerably as we move away from $a = 0$ (especially for $x > 0$ ), the tangent line, $T_{1} (x),$ is not a very good approximation for $x$ -values that satisfy $| x | > 0.5 .$ For example, $| e^{0.5} - T_{1} (0.5) | \approx 0.148721,$ so the linear approximation has an absolute error of more than $0.1$ at $x = 0.5 .$
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Using the strategy of finding a higher degree polynomial whose function and derivative values match at the selected point of tangency, we are able to find higher degree polynomials that much more effectively approximate $f (x) = e^{x}$ near $a = 0$ than the approximation generated by the tangent line. For example, using the degree $3$ approximation $T_{3} (x) = 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3},$ we see that $| f (x) - T_{3} (x) | < 0.01$ for all $x$ that satisfy $| x | < 0.6 .$
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It appears that the degree of the polynomial impacts the accuracy of the approximation of $f (x) = e^{x}$ in at least two ways: if we fix an $x$ -value, the higher the degree of the polynomial, the more accurate the approximation. In addition, raising the degree of the polynomial approximation appears to widen the interval on which the approximation is effective.
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Exercises 8.1.6 Exercises

1.

Let

f (x) = \cos (x)

and let

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} .

We find the values of

b_{0}, b_{1}

and

b_{2}

that make the functions

\cos (x)

and

T_{2} (x)

have the same height, slope, and concavity at

x = 0 .

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(a)

Complete the table below by computing derivatives of

T_{2} (x)

and

\cos (x)

and evaluating them at

x = 0

as directed.

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$f (x) =$	$\cos (x)$	$T_{2} (x) =$	$b_{0} + b_{1} x + b_{2} x^{2}$
$f^{'} (x) =$		$T_{2}^{'} (x) =$
$f^{″} (x) =$		$T_{2}^{″} (x) =$
$f (0) =$		$T_{2} (0) =$	$b_{0}$
$f^{'} (0) =$		$T_{2}^{'} (0) =$
$f^{″} (0) =$		$T_{2}^{″} (0) =$

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(b)

Now, since we want

T_{2} (x)

and

\cos (x)

to have the same height at

x = 0,

set

T_{2} (0) = f (0) .

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This implies

b_{0} =

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Since we want

T_{2} (x)

and

\cos (x)

to have the same slope at

x = 0,

set

T_{2}^{'} (0) = f^{'} (0) .

🔗

This implies

b_{1} =

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Finally, since we want

T_{2} (x)

and

\cos (x)

to have the same concavity at

x = 0,

set

T_{2}^{″} (0) = f^{″} (0) .

🔗

This implies

b_{2} =

🔗

Putting everything together, what is the resulting formula for

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} ?

🔗

T_{2} (x) =

🔗

2.

Let

f (x) = \ln (1 + x)

and let

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} .

We find the values of

b_{0}, b_{1}

and

b_{2}

that make the functions

\ln (1 + x)

and

T_{2} (x)

have the same height, slope, and concavity at

x = 0 .

🔗

(a)

Complete the table below by computing derivatives of

T_{2} (x)

and

\ln (1 + x)

and evaluating them at

x = 0

as directed.

🔗

$f (x) =$	$\ln (1 + x)$	$T_{2} (x) =$	$b_{0} + b_{1} x + b_{2} x^{2}$
$f^{'} (x) =$		$T_{2}^{'} (x) =$
$f^{″} (x) =$		$T_{2}^{″} (x) =$
$f (0) =$		$T_{2} (0) =$	$b_{0}$
$f^{'} (0) =$		$T_{2}^{'} (0) =$
$f^{″} (0) =$		$T_{2}^{″} (0) =$

🔗

(b)

Now, since we want

T_{2} (x)

and

\ln (1 + x)

to have the same height at

x = 0,

set

T_{2} (0) = f (0) .

🔗

This implies

b_{0} =

🔗

Since we want

T_{2} (x)

and

\ln (1 + x)

to have the same slope at

x = 0,

set

T_{2}^{'} (0) = f^{'} (0) .

🔗

This implies

b_{1} =

🔗

Finally, since we want

T_{2} (x)

and

\ln (1 + x)

to have the same concavity at

x = 0,

set

T_{2}^{″} (0) = f^{″} (0) .

🔗

This implies

b_{2} =

🔗

Putting everything together, what is the resulting formula for

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} ?

🔗

T_{2} (x) =

🔗

3.

Let

f (x) = e^{- 2 x}

and let

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} .

We find the values of

c_{0}, \dots, c_{3}

that make the functions

e^{- 2 x}

and

T_{3} (x)

have the same height, slope, concavity, and third derivative value at

x = 0 .

🔗

(a)

Complete the table below by computing derivatives of

T_{3} (x)

and

e^{- 2 x}

and evaluating them at

x = 0

as directed.

🔗

$f (x) =$	$e^{- 2 x}$	$T_{3} (x) =$	$c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3}$
$f^{'} (x) =$		$T_{3}^{'} (x) =$
$f^{″} (x) =$		$T_{3}^{″} (x) =$
$f^{‴} (x) =$		$T_{3}^{‴} (x) =$
$f (0) =$		$T_{3} (0) =$	$c_{0}$
$f^{'} (0) =$		$T_{3}^{'} (0) =$
$f^{″} (0) =$		$T_{3}^{″} (0) =$
$f^{‴} (0) =$		$T_{3}^{‴} (0) =$

🔗

(b)

Now, since we want

T_{3} (x)

and

e^{- 2 x}

to have the same height at

x = 0,

set

T_{3} (0) = f (0) .

🔗

This implies

c_{0} =

🔗

Since we want

T_{3} (x)

and

e^{- 2 x}

to have the same slope at

x = 0,

set

T_{3}^{'} (0) = f^{'} (0) .

🔗

This implies

c_{1} =

🔗

Since we want

T_{3} (x)

and

e^{- 2 x}

to have the same concavity at

x = 0,

set

T_{3}^{″} (0) = f^{″} (0) .

🔗

This implies

c_{2} =

🔗

Finally, since we want

T_{3} (x)

and

e^{- 2 x}

to have the same third derivative value at

x = 0,

set

T_{3}^{‴} (0) = f^{‴} (0) .

🔗

This implies

c_{3} =

🔗

Putting everything together, what is the resulting formula for

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} ?

🔗

T_{3} (x) =

🔗

4.

Let

f (x) = 3 x^{2} + x + 5

and let

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} .

We find the values of

b_{0}, b_{1}

and

b_{2}

that make the functions

f (x)

and

T_{2} (x)

have the same height, slope, and concavity at

x = 0 .

🔗

(a)

Complete the table below by computing derivatives of

T_{2} (x)

and

3 x^{2} + x + 5

and evaluating them at

x = 0

as directed.

🔗

$f (x) =$	$3 x^{2} + x + 5$	$T_{2} (x) =$	$b_{0} + b_{1} x + b_{2} x^{2}$
$f^{'} (x) =$		$T_{2}^{'} (x) =$
$f^{″} (x) =$		$T_{2}^{″} (x) =$
$f (0) =$		$T_{2} (0) =$	$b_{0}$
$f^{'} (0) =$		$T_{2}^{'} (0) =$
$f^{″} (0) =$		$T_{2}^{″} (0) =$

🔗

(b)

Now, since we want

T_{2} (x)

and

3 x^{2} + x + 5

to have the same height at

x = 0,

set

T_{2} (0) = f (0) .

🔗

This implies

b_{0} =

🔗

Since we want

T_{2} (x)

and

3 x^{2} + x + 5

to have the same slope at

x = 0,

set

T_{2}^{'} (0) = f^{'} (0) .

🔗

This implies

b_{1} =

🔗

Finally, since we want

T_{2} (x)

and

3 x^{2} + x + 5

to have the same concavity at

x = 0,

set

T_{2}^{″} (0) = f^{″} (0) .

🔗

This implies

b_{2} =

🔗

Putting everything together, what is the resulting formula for

T_{2} (x) = b_{0} + b_{1} x + b_{2} x^{2} ?

🔗

T_{2} (x) =

🔗

What does your answer mean about the quadratic approximation to a quadratic function?

🔗

5.

In the following problem, note that “second degree Taylor polynomial” is the same as the quadratic approximation we’ve been studying, and we’ll see this terminology again in the following section.

🔗

Suppose that

T_{2} (x) = c_{0} + c_{1} x + c_{2} x^{2}

is the second degree Taylor polynomial for the function

f

about

x = 0 .

What can you say about the signs of

c_{0},

c_{1},

c_{2}

f

has the graph given below?

🔗

(For each, enter $+$ if the term is positive, and $-$ if it is negative. Note that because this is essentially multiple choice problem it will not show which parts of your answer are correct or incorrect.)

🔗

c_{0}

🔗

c_{1}

🔗

c_{2}

🔗

6.

Throughout our work in Section 8.1, we have focused on approximating the function

f (x) = e^{x} .

In this exercise, we change the function of interest to

f (x) = \frac{1}{3} x^{3} + \frac{1}{4} x^{2} - 2 x - 1,

and consider the linear and quadratic approximations to

f

near

a = 0 .

🔗

Determine $f^{'} (x)$ and $f^{″} (x)$ and enter their formulas below.

$\begin{aligned} f (x) & = \frac{1}{3} x^{3} + \frac{1}{4} x^{2} - 2 x - 1 \\ f^{'} (x) & = \\ f^{″} (x) & = \end{aligned}$

🔗

🔗
Next, compute $f (0),$ $f^{'} (0),$ and $f^{″} (0)$ and enter those values below.

$\begin{aligned} f (0) & = \\ f^{'} (0) & = \\ f^{″} (0) & = \end{aligned}$

🔗

🔗
Use your work so far to determine the formula for $T_{1} (x),$ the tangent line approximation to $f (x)$ at $a = 0$ (which satisfies $T_{1} (0) = f (0)$ and $T_{1}^{'} (0) = f^{'} (0)$ ).
🔗

🔗
Let $T_{2} (x)$ be the quadratic approximation to $f (x)$ near $a = 0$ that satisfies $T_{2} (0) = f (0),$ $T_{2}^{'} (0) = f^{'} (0),$ and $T_{2}^{″} (0) = f^{″} (0) .$ You might start by letting $T_{2} (x) = c_{0} + c_{1} x + c_{2} x^{2},$ and creating an updated table like the one shown below.
🔗

$\begin{aligned} f (x) & = x^{3} / 3 + x^{2} / 4 - 2 x - 1 \\ f^{'} (x) & = \\ f^{″} (x) & = \\ f (0) & = \\ f^{'} (0) & = \\ f^{″} (0) & = \end{aligned}$

🔗

$\begin{aligned} T_{2} (x) & = c_{0} + c_{1} x + c_{2} x^{2} \\ T_{2}^{'} (x) & = c_{1} + 2 c_{2} x \\ T_{2}^{″} (x) & = \\ T_{2} (0) & = \\ T_{2}^{'} (0) & = \\ T_{2}^{″} (0) & = \end{aligned}$

🔗

Use your work in in the table above to find the formula for $T_{2} (x)$ that satisfies $T_{2} (0) = f (0),$ $T_{2}^{'} (0) = f^{'} (0),$ and $T_{2}^{″} (0) = f^{″} (0) .$
🔗

🔗
Plot $f (x),$ $T_{1} (x),$ and $T_{2} (x)$ on the same axes, centered at $a = 0 .$ What do you notice?
🔗

🔗

🔗

7.

In this exercise, we extend our work in Exercise 8.1.6.6. We continue to consider the function

f (x) = \frac{1}{3} x^{3} + \frac{1}{4} x^{2} - 2 x - 1,

but now build the cubic (degree

3

) approximation to

f

near

a = 0 .

🔗

Let $T_{3} (x) = k_{0} + k_{1} x + k_{2} x^{2} + k_{3} x^{3}$ and determine $T_{3}^{'} (x),$ $T_{3}^{″} (x),$ and $T_{3}^{‴} (x) .$ In addition, determine $f^{‴} (x) .$ Record your results below, along with the values of each of these functions at $a = 0 .$
🔗

$\begin{aligned} f (x) & = x^{3} / 3 + x^{2} / 4 - 2 x - 1 \\ f^{'} (x) & = \\ f^{″} (x) & = \\ f^{‴} (x) & = \\ f (0) & = \\ f^{'} (0) & = \\ f^{″} (0) & = \\ f^{‴} (0) & = \end{aligned}$

🔗

$\begin{aligned} T_{3} (x) & = k_{0} + k_{1} x + k_{2} x^{2} + k_{3} x^{3} \\ T_{3}^{'} (x) & = \\ T_{3}^{″} (x) & = \\ T_{3}^{‴} (x) & = \\ T_{3} (0) & = \\ T_{3}^{'} (0) & = \\ T_{3}^{″} (0) & = \\ T_{3}^{‴} (0) & = \end{aligned}$

🔗

🔗
Use your work in above to find the formula for $T_{3} (x)$ that satisfies $T_{3} (0) = f (0),$ $T_{3}^{'} (0) = f^{'} (0),$ $T_{3}^{″} (0) = f^{″} (0),$ and $T_{3}^{‴} (0) = f^{‴} (0) .$
🔗

🔗
Plot $f (x),$ $T_{1} (x),$ $T_{2} (x),$ and $T_{3} (x)$ on the same axes, centered at $a = 0 .$ What do you notice?
🔗

🔗
What do you expect the degree $4$ approximation to $f (x),$ $T_{4} (x)$ to be? Why?
🔗

🔗
What do you expect will happen if we find the degree $6$ approximation to any function $f (x)$ that is itself a degree $6$ polynomial?
🔗

🔗

🔗

You have attempted 1 of 7 activities on this page.

🔗

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Section 8.1 Extending local linearization

Motivating Questions

Subsection 8.1.1 Introduction

Preview Activity 8.1.1.

(a)

(b)

(c)

(d)



Subsection 8.1.2 Finding a quadratic approximation

The quadratic approximation of f(x)=ex.



Activity 8.1.2.

(a)

(b)

(c)

(d)

(e)

(f)

Subsection 8.1.3 Over and over again



Activity 8.1.3.

(a)

(b)

(c)

(d)

Subsection 8.1.4 As the degree of the approximation increases



Activity 8.1.4.

(a)

(b)

(c)

(d)

(e)

(f)

Subsection 8.1.5 Summary

Exercises 8.1.6 Exercises

1.

(a)

(b)

2.

(a)

(b)

3.

(a)

(b)

4.

(a)

(b)

5.

6.

7.

The quadratic approximation of $f (x) = e^{x}$ .