Last updated on August 5th, 2025
A calculator is a tool designed to perform both basic arithmetic operations and advanced calculations, such as those involving calculus. It is especially helpful for completing mathematical school projects or exploring complex mathematical concepts. In this topic, we will discuss the Linear Approximation Calculator.
The Linear Approximation Calculator is a tool designed for estimating the value of a function near a given point using its tangent line.
Linear approximation is a method used in calculus to approximate the value of a function at a certain point by using the function's derivative.
This method uses the concept that the tangent line to the curve of a function at a given point can serve as an approximation of the function near that point.
To calculate the linear approximation using the calculator, we need to follow the steps below -
Step 1: Input: Enter the function and the point of approximation.
Step 2: Click: Calculate Approximation.
By doing so, the function we have given as input will get processed.
Step 3: You will see the linear approximation of the function at the specified point in the output column.
Mentioned below are some tips to help you get the right answer using the Linear Approximation Calculator.
Know the formula: The formula for linear approximation is \( L(x) = f(a) + f'(a)(x-a) \), where \( f(a) \) is the function value at \( a \) and \( f'(a) \) is the derivative at \( a \).
Use the Right Units: Make sure the function and the point are in compatible units, if applicable.
Enter correct Values: When entering the function and the point, make sure the values are accurate.
Small mistakes can lead to big differences, especially with complex functions.
Calculators mostly help us with quick solutions.
For calculating complex math questions, students must know the intricate features of a calculator.
Given below are some common mistakes and solutions to tackle these mistakes.
Help Sarah approximate the value of \( f(x) = \sin(x) \) at \( x = \pi/4 \).
We find the linear approximation of the function to be \( L(x) = \sqrt{2}/2 + \sqrt{2}/2(x - \pi/4) \).
To find the linear approximation, we use the formula: \( L(x) = f(a) + f'(a)(x-a) \).
Here, the value of \( a \) is \( \pi/4 \).
The function value \( f(\pi/4) = \sin(\pi/4) = \sqrt{2}/2 \).
The derivative \( f'(x) = \cos(x) \), so \( f'(\pi/4) = \sqrt{2}/2 \).
Therefore, \( L(x) = \sqrt{2}/2 + \sqrt{2}/2(x - \pi/4) \).
Approximate the value of \( f(x) = \ln(x) \) at \( x = 1 \).
The linear approximation is \( L(x) = (x - 1) \).
To find the linear approximation, we use the formula: \( L(x) = f(a) + f'(a)(x-a) \).
Since the value of \( a \) is 1, The function value \( f(1) = \ln(1) = 0 \).
The derivative \( f'(x) = 1/x \), so \( f'(1) = 1 \).
Thus, \( L(x) = 0 + 1(x - 1) = x - 1 \).
Find the linear approximation of \( f(x) = e^x \) at \( x = 0 \).
We will get the approximation as \( L(x) = 1 + x \).
For the linear approximation, we use the formula: \( L(x) = f(a) + f'(a)(x-a) \). At \( a = 0 \), The function value \( f(0) = e^0 = 1 \).
The derivative \( f'(x) = e^x \), so \( f'(0) = 1 \).
Thus, \( L(x) = 1 + 1(x - 0) = 1 + x \).
The function \( f(x) = x^2 \) needs approximation at \( x = 2 \). Find the linear approximation.
We find the linear approximation to be \( L(x) = 4 + 4(x - 2) \).
Using the linear approximation formula: \( L(x) = f(a) + f'(a)(x-a) \).
At \( a = 2 \), The function value \( f(2) = 2^2 = 4 \).
The derivative \( f'(x) = 2x \), so \( f'(2) = 4 \). Thus, \( L(x) = 4 + 4(x - 2) \).
Approximate \( f(x) = \sqrt{x} \) at \( x = 4 \).
The linear approximation is \( L(x) = 2 + \frac{1}{4}(x - 4) \).
Using the formula \( L(x) = f(a) + f'(a)(x-a) \): For \( a = 4 \), The function value \( f(4) = \sqrt{4} = 2 \).
The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \), so \( f'(4) = \frac{1}{4} \).
Thus, \( L(x) = 2 + \frac{1}{4}(x - 4) \).
Seyed Ali Fathima S a math expert with nearly 5 years of experience as a math teacher. From an engineer to a math teacher, shows her passion for math and teaching. She is a calculator queen, who loves tables and she turns tables to puzzles and songs.
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