Linear Approximation

Linear approximation is a method used to estimate the value of a function near a given point by using the equation of its tangent line at that point.

Here, a differentiable function behaves like a straight line when examined on a sufficiently small scale. By using the first-degree Taylor expansion, we approximate the function with its tangent line, which has the same slope as the function at a given point.

Mathematically, the linear approximation of a function f(x) near x = a is given by:

L(x) = f(a) + f’(a)(x – a)

Here,

• L(x) is the linear approximation
• f(a) is the function value at x = a
• f’(a) is the derivative of f(x) at x = a
• (x – a) represents the small change from the point of approximation

This approximation is particularly effective when x is close to a, as the actual function and its tangent line remain nearly indistinguishable in a local neighborhood. 

Steps

Now, let us estimate ${\sqrt{4.1}}$ without directly calculating the square root.

Step 1: Defining the Function

The function can be expressed as:

f(x) = ${\sqrt{x}}$

Since square roots can be difficult to calculate mentally, we will approximate it using the tangent line at a nearby simple value.

Step 2: Choosing a Reference Point x = a Close to 4.1

To approximate ${\sqrt{4.1}}$, we select a value close to 4. Since ${\sqrt{4}}$ is easy to calculate, so we choose a = 4

f(4) = ${\sqrt{4}}$ = 2

Step 3: Finding the Derivative of f(x) 

Since linear approximation relies on the slope of the tangent line, we first compute the derivative of f(x) = ${\sqrt{x}}$:

f’(x) = ${\dfrac{1}{2\sqrt{x}}}$

Step 4: Evaluating the Derivative at x = 4 

Now, substituting x = 4 into the derivative formula, we get

f’(4) = ${\dfrac{1}{2\sqrt{4}}}$ = ${\dfrac{1}{2\left( 2\right) }}$ = ${\dfrac{1}{4}}$

This means that at x = 4, the slope of the tangent line is ${\dfrac{1}{4}}$

Step 5: Applying the Linear Approximation Formula

The general formula for linear approximation is:

L(x) = f(a) + f’(a)(x – a)

Substituting the known values:

f(4) = 2

f’(4) = ${\dfrac{1}{4}}$

a = 4

Now, L(x) = ${2+\dfrac{1}{4}\left( x-4\right)}$

Step 6: Approximating ${\sqrt{4.1}}$

Now, substituting x = 4.1 into the linear approximation formula, we get

L(4.1) = ${2+\dfrac{1}{4}\left( 4.1-4\right)}$

⇒ L(4.1) = ${2+\dfrac{0.1}{4}}$

⇒ L(4.1) = ${2+0.025}$

⇒ L(4.1) = ${2.025}$

Thus, the approximation for ${\sqrt{4.1}}$ is 2.025

Step 7: Calculating the Error

To check the accuracy of our approximation, we compare it with the actual value obtained using a calculator:

${\sqrt{4.1}}$ ≈ 2.0248

The absolute error is (2.025 – 2.0248) = 0.0002

The percentage error is (0.0002 × 100)% = 0.02%

Since the error is very small, this confirms that linear approximation works well for small changes in x.

Solved Examples