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Introduction to Differential Equations - At A Glance:

Tangent line approximation can also be called local linearization, linear approximation, and probably a bunch of other names. The important thing is that you're using a line to approximate a curve.

Here's a reminder of how tangent line approximation works:

Suppose we know the value of f at a particular value of x:

and we want to know the value of f at a nearby value of x:

We draw the tangent line to f at the point we know:

Then we find the value on the tangent line at the nearby value of x:

This tangent line value is close to the value we actually wanted.

Using the language from the last section, let's talk this through again.

We know yold:

We want to know the value of f if we change x by Δ x:

We draw the tangent line to f at the point we know:

and we use the tangent line to find ynew, which is close to the value we actually wanted.

Sample Problem

Another kind of problem you may run into is the kind where you're given a formula for the function f and asked to use a tangent line to approximate f at some particular value x*.

To do this, pick some x = a close to x*. Choose a so that f (a) and f '(a) are easy to calculate. Then we have yold = f (a), Δ x = x* - a, and slope = f '(a).

From there you know how to calculate ynew, which is the approximation you want.

Example 1

Let f (x) = 4x2 + 3.

(a) Find the tangent line to f at x = 3.
(b) Use the tangent line from part (a) to approximate f at 3.001.


Example 2

If f (1) = 2 and f ' (1) = 3, estimate f (1.2).


Example 3

The picture below shows the tangent line to the function f at x = 0. Estimate f (-0.1).


Example 4

The picture below shows a tangent line to f at x = a. What is f '(a)?


Example 5

Let f (x) = cos2 x. Use a tangent line approximation to estimate f (0.1).


Example 6

Let f (x) = x5 + 3x4 - 2x2 - x. Use a tangent line approximation to estimate f (.9).


Example 7

Can a tangent line approximation ever produce the exact value of the function? Why or why not?


Example 8

The function y = f (x) is a solution to the IVP

Approximate f (1.01).


Exercise 1

For the function,

(a) Find the tangent line to the function at the specified value of x.

(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).

f (x) = x3 + 2x + 1, x = 1, Δ x = .2

Exercise 2

For the function,

(a) Find the tangent line to the function at the specified value of x.

(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).

Exercise 3

For the function,

(a) Find the tangent line to the function at the specified value of x.

(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).

f (x) = log2 x, x = 8, Δ x = .01

Exercise 4

For the function,

(a) Find the tangent line to the function at the specified value of x.

(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).

f (x) = 4x + ex, x = 0, Δ x = 0.05

Exercise 5

For the function,

(a) Find the tangent line to the function at the specified value of x.

(b) Use the tangent line from (a) to estimate the value of the function at (x + Δ x).

Exercise 6

If f (3) = 4.2 and f '(3) = 0.15, estimate f (3.1).

Exercise 7

If and , estimate .

Exercise 8

The picture below shows the tangent line to f at -1. Estimate f (-.9).

Exercise 9

The picture below shows the tangent line to f at x = 3. Find f '(3).

Exercise 10

If f (2) = 7 and a tangent line approximation at x = 2 estimates f (2.1) to be 7.2, what is f '(2)?

Exercise 11

Let f (x) = 3x2 - 4x + 1. Use a tangent line approximation to estimate f (2.01).

Exercise 12

Let f (x) = 7x2 - 3x + 4. Use a tangent line approximation to estimate f (2.99).

Exercise 13

Let f (x) = ex. Use a tangent line approximation to estimate f (0.05).

Exercise 14

Let f (x) = 2x. Use a tangent line approximation to estimate f (3.02).

Exercise 15

Let f (x) = sin (5x). Use a tangent line approximation to estimate f (3).

Exercise 16

The function y = f (x) is a solution to the IVP

Estimate f (0.5).

Exercise 17

The function y = f (x) is a solution to the IVP

Estimate f (-½).

Exercise 18

The function y = f (x) is a solution to the IVP

Estimate f (2.01).

Exercise 19

The function y = f (x) is a solution to the IVP

Approximate f (6.95).

Exercise 20

The function y = f (x) is a solution to the IVP

Approximate f (3.01).

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