Linear Approximation Formula
The concept behind the linear approximation formula is the equation of a tangent line. We know that the slope of the tangent that is drawn to a curve y = f(x) at x = a is its derivative at that point. i.e., the slope of the tangent line is f'(a). Thus, the linear approximation formula is an application of derivatives. Let us learn more about this formula in the upcoming sections.
What Is Linear Approximation Formula?
As we discussed in the previous section, the linear approximation formula is nothing but the equation of a tangent line. Let us find the equation of a tangent line that is drawn to the curve y = f(x) at the point x = a (or) (a, f(a)). As we know the slope of this tangent is the derivative f '(a), its equation using the pointslope form is,
y  f(a) = f '(a) (x  a)
(or) y = f(a) + f '(a) (x  a)
This is exactly the equation that is used as the linear approximation formula, and we denote this as L(x) (where "L" stands for linear approximation). Thus, the linear approximation formula is:
L(x) = f(a) + f '(a) (x  a)
where,
 L(x) is the linear approximation of f(x) at x = a.
 f '(a) is the derivative of f(x) at x = a.
L(x) is used to approximate the values of f(x) at the values of x that are very close to "a". Let us see the applications of the linear approximation formula in the following solved examples section.
Solved Examples Using Linear Approximation Formula

Example 1: Find the linear approximation of f(x) = √x at x = 4. Using this, find the approximate value of √4.04.
Solution:
The given function is, f(x) = √x.
We have to find the linear approximation of f(x) at a = 4.
So f(a) = √4 = 2.
The derivative of f(x) is,
f ' (x) = d/dx (√x) = d/dx (x^{1/2}) = 1/2 . x ^{1/2} = 1/ (2√x)
f ' (a) = f ' (4) = 1/ (2√4) = 1/4.
The linear approximation formula of f(x) is,
L(x) = f(a) + f '(a) (x  a)
L(x) = 2 + 1/4 (x  4)
L(x) = 2 + (1/4)x  1
L(x) = (1/4)x + 1
Using this,
√4.04 ≈ L(4.04) = (1/4) (4.04) + 1 = 2.01
Answer: The linear approximation of f(x) is, L(x) = (1/4)x + 1 and √4.04 ≈ 2.01.

Example 2 : Use the linear approximation formula to find the approximate value of \(\sqrt[3]{27.05}\). Round your answer to 4 decimals.
Solution:
Let us assume that f(x) = \(\sqrt[3]{x}\).
Since 27.05 is very close to 27, let us find the linear approximation of f(x) at x = 27. Thus, a = 27.
f(a) = \(\sqrt[3]{27}\) = 3.
The derivative of f(x) is,
f ' (x) = d/dx (\(\sqrt[3]{x}\)) = d/dx (x^{1/3}) = 1/3 . x ^{2/3}
f ' (a) = f ' (27) = 1/3 . (27) ^{2/3} = 1/27
The linear approximation formula of f(x) is,
L(x) = f(a) + f '(a) (x  a)
L(x) = 3 + 1/27 (x  27)
L(x) = 3 + (1/27)x  1
L(x) = (1/27)x + 2
Using this,
\(\sqrt[3]{27.05}\) ≈ L(27.05) = (1/27)(27.05) + 2 = 3.0019
Answer: \(\sqrt[3]{27.05}\) ≈ 3.0019