The equation to tangent line at the point (a,f(a)) is given by
y = f(a) + f’(a)(x  a).
The linear approximation to the function when x is close to a is given by
f(x) $\approx$ f(a) + f’(a)(x  a).
As we use the tangent line equation for approximation, linear approximation is also called
tangent line approximation. The Linear function, which represents the equation to the tangent line
L(x) = f(a) + f’(a) (x  a) is called the
Linearization of ‘f’ at ‘a’.
Linear approximation formula is therefore given by f(x) $\approx$ f(a) + f’(a)(x  a)
The linear approximation f(x) $\approx$ f(a) + f'(a) (x  a) varies as the value of a varies. The values of x obtained using Linear approximation will deviate more from the actual function value as x moves further from a. The Linearization can not be considered good when the difference x  a is not really small.
Let us consider the linear approximation for $f(x) = \sqrt{x + 2}$ around $x = 2$
$f(x) = f(1) + f'(1) (x  1) = 2 + $
$\frac{1}{4}$$(x  2)$ =
$\frac{3}{2}$ +
$\frac{x}{4}$The graph for $y = \sqrt{x + 2}$ and the tangent line $Y$ =
$\frac{3}{2} + \frac{x}{4}$ are shown below.
The following table gives the approximated value using the linearization and the actual values for various values of x
x

F(x)

Actual value
 Linear approximation
 Error

0 
$\sqrt{2}$

1.4142135623Ã¢â‚¬Â¦ 
1.5 
> 0.05 
0.5 
$\sqrt{2.5}$ 
1.5811388300... 
1.625 
> 0.001 
1.9 
$\sqrt{3.9}$ 
1.9748417658... 
1.975 
< 0.0002 
1.98 
$\sqrt{3.98}$ 
1.9949937343... 
1.995 
< 0.00001 
2 
$\sqrt{4}$ 
2

2

NIL 
2.02 
$\sqrt{4.02}$ 
2.0049937655Ã¢â‚¬Â¦ 
2.005

< 0.00001 
2.5 
$\sqrt{4.5}$ 
2.12132034355.. 
2.125 
> 0.001 
3 
$\sqrt{5}$ 
2.23606797749.. 
2.25 
> 0.01 
10 
$\sqrt{12}$ 
3.46410161513.. 
4 
> 0.5 
100 
$\sqrt{102}$ 
10.0995049383..  26.5  > 16 
We observe from the graph and the table, the tangent function at x = 2, used as a Linear approximation, is close enough to the actual values, only when x is close to 2. As the value x deviates from x = 2 on both sides, that is both when x > 2 and x < 2, the Linearization value is not a good approximation to the actual value. Hence any Linear approximation used at a given value of x is called the Local Linear approximation. Thus, t Linear approximation which is an application of derivative used as a local estimate brings in to light the easy utilization of the derivative of a function.
Linear Approximation Applications
Imagine you need to estimate the value of (1.0002)
^{50}. Possibly you are calculating the return of an investment. You realize that you have forgotten to take the calculator. Linear approximation comes to your rescue now.
Consider the function f(x) = (1+x)
^{50} The Linear approximation for the function around x = 0 is
f(x) = f(0) + 50(1)
^{49}(x) = 1 + 50(x)
(1.0002)
^{50} = (1+ 0.0002)
^{50} = 1 + 50(0.0002) = 1 + 0.01 = 1.01
Linear approximation is handy even when the function is not defined, but the observations are given as a table of values. Linear approximation here, is used to interpolate and estimate f'(a) from the table values. This is done without actually finding the trend line for the data set.
Linear approximation finds application in Physics. When the deflection $\theta$ is very small, which means close to zero, sin$\theta$ is approximated to $\theta$ and $\cos \theta$ to 1 using linear approximation. Euler uses a mix of Linearization and direction fields in his numerical approach to solve a differential equation.
To find the linear approximation of a given function, the necessary condition is that the function is differentiable. If its not then we don't able to find the approximation of that function. If f(x) is a given function which is differentiable also then there is a tangent line at any point on the graph of the function. Let f(x) is differentiable at the point a, then the tangent to the curve at a point (a,f(a)) on a graph is given by
y  f(a) = f'(a) (x  a) where, f(a) is the value of the function at the point a.
The tangent ling is also the graph of the function which is linear, so we say it as L
_{a}(let), since its a linear relation. The function L
_{a} is the beat linear approximation to the function f near a.
L
_{a} = f(a) + f'(a) (x  a)
The linear function L
_{a} is an approximation to the given function f(x) as:
f(x) $\approx$ L
_{a}(x) = f(a) + f'(a) (x  a)
This is called linear approximation or tangent line approximation of a given function f(x).
Given below are some examples on linear approximation.
Solved Examples
Question 1: Find the linear approximation for x tends to 0 where f(x) = tan x.
Solution:
We have to first find out f '(0)
We know f(x) = tan x
f'(x) = sec^{2}x
So, 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)
= 0 + x = x
The above result means that tan x $\approx$ x for x close to 0
Question 2: Find the linear approximation for x tends to 1 where f(x) = ln x.
Solution:
We have to find out f'(1) first.
We know f(x) = ln x
f'(x) = $\frac{1}{x}$
f'(1) = 1
Hence the linear approximation f_{l}(x) is given by,
f_{l}(x) = ln 1 + f'(1) (x  1)
= x  1
This means that
ln x $\approx$ x  1 for x close to 1.
Question 3: Find the linear approximation of f(x) = e
^{x}, for x close to 0.
Solution:
We have to find f'(0)
We have f(x) = e^{x}
f'(x) = e^{x}
f'(0) = 1
Hence the linear approximation f_{l}(x) is given by,
f_{l}(x) = e^{0} + f '(0) (x  0)
= 1 + x
It means that
e^{x} $\approx$ x + 1 for x close to 0.
Question 4: Consider the function f(x) = $\sqrt{1 + x}$. The linear approximation for values of x close to x = 0 is given by f(x) $\approx$ f(0) + f’(0)x.
Solution:
f(0) = $\sqrt{1 + 0}$ = 1
f’(x) = $\frac{1}{2 \sqrt{1 + x}}$
Hence f’(0) = $\frac{1}{2 \sqrt{1 + 0}}$ = $\frac{1}{2}$
The required Linear approximation around x = 0 is hence
f(x) = 1 + $\frac{1}{2}$x
In mathematics, if a function defined in terms of piecewise
is called piecewise function. And, if all pieces are in linear form then that function is known as piecewise linear function.
A piecewise linear approximation of the given system obtained by solving the right side of the grid defined above. Let f is a continuous function on R and f is linear on [a
_{i},a
_{i+1}] where i = 1,2 ,.., k and a
_{1} < a
_{2 }< a
_{3 }< ..... < a
_{k} all are real numbers and a
_{i}'s are called the break points.These piecewise linear function are used to approximate nonlinear functions. The simple approach of constructing the piecewise linear approximation is to make a smooth function. So, we can calculate a function at a number of grid points and after that use linear interpolation between these given points to calculate the approximate.
The purpose of this is to show how and why.
 We can write every piecewise linear function as a sum of a constant + linear term + a sum of absolute value, like $f(x)= c + a_{0}+ \sum_{i=1}^{k} a_{i}\left  x  b_{i} \right $ where c_{i}'s an explicit expression for each of a_{j}'s(including a0 in the term of c_{i}'s).
 The term c_{i}'s gives the necessary and sufficient condition for f to be convex.
 The term a_{j}'s gives the necessary and sufficient condition for f to be convex.
Sometimes the concept behind the linear approximation are expressed in
the terminology and notations of the differentials. If we have y = f(x) is a differentiable function then differentiable dx is an
independent variable. It means dx can be given the value of real number. So, we can easily express dy in terms of dx as dy = f'(x) dx, where x
is taken any value from the domain of f and dx gives a specify value.
We can determine the value of dy with the help of x and dx hence dy
depends on x and dx.
If we draw the graph of f(x), then let A(x,
f(x)) and B(x + $\triangle$x , f(x + $\triangle$x)) be the two points on
the graph and let
$dx = \triangle x$. Then, $y = \triangle y = f(x + \triangle x)  f(x)$
Since, the derivative of the function f(x) i.e. f'(x) is the slope of
the tangent line AC. The distance from D to C is f'(x) dx = dy. Here, dy
represents the rises or falls in the tangent line and $\triangle y$
represents the curve y = f(x) falls or rises when x made a small change
dx.
Let dx = x  a then x = a + dx. Then, the formula for linear approximation f(x) $\approx$ f(a) + f'(a)(x  a) can be written in terms of differential as f(a + dx) = f(a) + dy
$\triangle y = dy$ becomes better if $\triangle x$ is smaller