# Linear Approximation of Functions

Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. Examples with detailed solutions on linear approximations are presented.

## Linear Approximations to Functions

A possible linear approximation f_{l} to function f at x = a may be obtained using the equation of the tangent line to the graph of f at x = a as shown in the graph below.

f_{l}(x) = f(a) + f '(a) (x - a)

For values of x closer to x = a, we expect f(x) and f_{l}(x) to have close values. Since f_{l}(x) is a linear function we have a linear approximation of function f.

This approximation may be used to linearize non algebraic functions such as sine, cosine, log, exponential and many other functions in order to make their computation easier. Examples are presented below.

### Example 1

Find the linear approximation of f(x) = tan x, for x close to 0.

### Solution to Example 1

We first compute f '(0)

f '(x) = sec ^{2} x

f '(0) = sec ^{2} (0) = 1

Hence the linear approximation f_{l}(x) is given by

f_{l}(x) = f(0) + f '(0) (x - 0) = x

The above result means that tan x ≈ x for x close to 0 when x is in **RADIANS**.

Put your calculator to **RADIANS** and calculate tan x for the following values of x.

x = 0 , x = 0.001 , x = 0.01, x = 0.1, x = 0.2, x = 0.3 and x = 0.5

Note compare tan x and x. Conclusion.

### Example 2

Find the linear approximation of f(x) = ln x, for x close to 1.

### Solution to Example 2

We first compute f '(1)

f '(x) = 1 / x

f '(1) = 1

Hence the linear approximation f_{l}(x) is given by

f_{l}(x) = ln 1 + f '(1) (x - 1) = x - 1

The above result means that

ln x ≈ x - 1 for x close to 1.

Use your calculator to calculate ln x and x - 1 for

x = 1 , x = 1.001 , x = 1.01, x = 1.1, x = 1.5

Note compare ln x and x - 1. Conclusion.

### Example 3

Find the linear approximation of f(x) = e^{x}, for x close to 0.

### Solution to Example 3

f '(0) = 1

Hence the linear approximation f_{l}(x) is given by

f_{l}(x) = e^{0} + f '(0) (x - 0) = 1 + x

Use your calculator to calculate e^{x} and 1 + x for

x = 0 , x = 0.001 , x = 0.01, x = 0.1 and x = 0.5

and compare.
Linear approximation is one of the simplest approximations to transcendental functions that cannot be expressed algebraically. However there are other more powerful methods that give better algebraic approximations to these functions.

### More on applications of differentiation

applications of differentiation