5- What is the Linear Approximation method?

Last Updated on January 19, 2026 by Maged kamel

What is the Linear Approximation method?

What is Linear approximation?

Linear approximation, or linearization, is a method for approximating the value of a function at a given point. There is a reason linear approximation is useful: it can be difficult to find the value of a function at a particular point.

We have a function like that. We know the value of x of a one-point, which is called a. We know xa, which is the x-distance, and we know the coordinate value, which is f(x=a).

Suppose we have another point that is considered close to point A. We can assume that we have a line tangent to that curve at point a.

We can treat the function as a linear one at any point; we get the y-value from that line.

The linear approximation equation.

So the linear approximation is y = the value of any point we need, starting from a given point xa and its value f(a). We make this line with slope, then go to another point, say c. We get the value of xc by substituting. We get the function value of x_c. 

Linear approximation is a method for finding the y-value of point b, given the x-value of another point, a, and its y-value. The linear approximation value L (x=xb) is equal to the function value. at a+( the slope of the tangent) *( the difference. between(the new Point x_b – the old Point x_a).

Solved problem#1: Find the linear approximation of y as f(x)=sin (x) at x0=0.

Our solved problem: Find the linear approximation of y as f(x)=sin(x) at x0=0. We put the equation in the general form by writing. L(x)= approximately=f(x=a)+f'(a) *(xb-xa), the difference between point b and point a, where point a is the starting point for which we know the value of the Y as well as the f’ value.

To develop L(x) for f(x)=sin(x), we get the f'(x) expression, which is cos(x), the f(x) value at x0=0 will be equal to=0, while the f'(x=0)=cos(0)=1. We consider (xb-x0)=x, the formula L(x)=0+(x-0)*1=X.

For part b, it is required to find the local linear approximation value of the function at x=2 degrees. L(x=2)=2 degrees; we convert that to rad and get 0.0349066. The exact value of f(x=2)= sin(2)=0.0348995. Please refer to the following slide for more details.

The following graph shows the f(x), which is sin(x), and the line of L(x) that passes through the x0=0 point.

To estimate the absolute value of the error, we deduct the value of L(x) from the actual value. For our solved problem #1, we get a value of 7.6*10^-6.

Solved problem#2: find the linear approximation of y as f(x)=x^2 at 2.50.

As an application, we can obtain a linear approximation of any function and find the power, square root, or cube root of a number.

Our solved problem: find the linear approximation of y as f(x)=x^2 at 2.50. We put the equation in the general form by writing. L(x)= approximately=f(x=a)+f'(a) *(xb-xa), the difference between point b and point a, where point a is the starting point for which we know the value of the Y as well as the f’ value.

Our next question is, what is the nearest Point to 2.50? The nearest point to 2.50 is 2, for which we can get the value of y, as well as the value of the slope.
 1- We need to estimate the function value at the starting point a, which is equal to (2.0)^2=4.00.
2- We need to estimate the slope value, the x^2 to be differentiated as y’=2*x
3- Get the value of y’ at x=2, it will be=2*(2)=4.0.

4-We can apply the expression and get L(x=2.50) =f(2.0)+(y’*(x_b-x_a), a is the starting point where x=2.0, while xb is the point for which we want to estimate the Y value, which is equal to 2.50.
5-The value of L(x=2.50)=4+((2)*(2)*(2.50-2.0))=6.00.
6-The exact value= the approximated value + error, apply in the equation of y=x^2, the exact value is (2.50)^2=6.25, the approximated value=6.00. 7- The error value=6.25-6.0=0.25.

The following graph shows the f(x), which is x^2, and the line of L(x) that passes through the x0=2 point.

The PDF data for this post and the following post can be reviewed or downloaded through the following document.

The next post is about the Practice problems with linear approximation.

Here is a useful link to the Linear approximation calculator-e math help.