# Linear Approximation

If the function $$y = f\left( x \right)$$ is differentiable at a point $$a$$, then the increment of this function when the independent variable changes by $$\Delta x$$ is given by

$\Delta y = A\Delta x + \omicron\left( {\Delta x} \right),$

where the first term $$A\Delta x$$ is the differential of function, and the second term has a higher order of smallness with respect to $$\Delta x.$$ The differential of function is denoted by $$dy$$ and is linked with the derivative at the point $$a$$ as follows:

$dy = A\Delta x = f'\left( {a} \right)\Delta x.$

Thus, the increment of the function $$\Delta y$$ can be written as

$\Delta y = dy + \omicron\left( {\Delta x} \right) = f'\left( {a} \right)\Delta x + \omicron\left( {\Delta x} \right).$

For sufficiently small increments of the independent variable $$\Delta x$$, one can neglect the "nonlinear" additive term $$\omicron\left( {\Delta x} \right).$$ In this case, the following approximate equality is valid:

$\Delta y \approx dy = f'\left( {a} \right)\Delta x.$

Note that the absolute error of the approximation, i.e. the difference $$\Delta y - dy$$ tends to zero as $$\Delta x \to 0:$$

$\require{cancel} \lim\limits_{\Delta x \to 0} \left( {\Delta y - dy} \right) = \lim\limits_{\Delta x \to 0} \left[ {\cancel{dy} + \omicron\left( {\Delta x} \right) - \cancel{dy}} \right] = \lim\limits_{\Delta x \to 0} \omicron\left( {\Delta x} \right) = 0.$

Moreover, the relative error also tends to zero as $$\Delta x \to 0:$$

$\lim\limits_{\Delta x \to 0} \frac{{\Delta y - dy}}{{dy}} = \lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{f'\left( {a} \right)\Delta x}} = \frac{1}{{f'\left( {a} \right)}}\lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}} = 0,$

since $$\omicron\left( {\Delta x} \right)$$ corresponds to the term of the second and higher order of smallness with respect to $$\Delta x.$$

Thus, we can use the following formula for approximate calculations:

$f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right).$

where the function $$L\left( x \right)$$ is called the linear approximation or linearization of $$f\left( x \right)$$ at $$x = a.$$

Linear approximation is a good way to approximate values of $$f\left( x \right)$$ as long as you stay close to the point $$x = a,$$ but the farther you get from $$x = a,$$ the worse your approximation.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Let $$f\left( x \right)$$ be a differentiable function such that $$f\left( 3 \right) = 12,$$ $$f^\prime\left( 3 \right) = - 2.$$ Estimate the value of $$f\left( {3.5} \right)$$ using the local approximation at $$a = 3.$$

### Example 2

Let $$g\left( x \right)$$ be a differentiable function such that $$g\left( { - 2} \right) = 0,$$ $$g^\prime\left( { - 2} \right) = - 5.$$ Find the value of $$g\left( { - 1.8} \right)$$ using the local approximation at $$a = - 2.$$

### Example 3

Find an approximate value for $$\sqrt[3]{{30}}.$$

### Example 4

Estimate $$\sqrt[3]{9}$$ using a linear approximation at $$a = 8.$$

### Example 5

Calculate an approximate value of $$\sqrt {50}.$$

### Example 6

Estimate $$\sqrt {3.9}$$ using a linear approximation at $$a = 4.$$

### Example 7

Calculate an approximate value for $$\sqrt[4]{{0,025}}.$$

### Example 8

Find the linearization of the function $f\left( x \right) = \sqrt[3]{{{x^2}}}$ at $$a = 27.$$

### Example 9

Calculate $${\left( {8,2} \right)^{\frac{2}{3}}}.$$

### Example 10

Derive the approximate formula ${\left( {1 + \alpha } \right)^n} \approx 1 + n\alpha .$ Calculate the approximate value for $$\sqrt {1,02} .$$

### Example 1.

Let $$f\left( x \right)$$ be a differentiable function such that $$f\left( 3 \right) = 12,$$ $$f^\prime\left( 3 \right) = - 2.$$ Estimate the value of $$f\left( {3.5} \right)$$ using the local approximation at $$a = 3.$$

Solution.

The linear approximation is given by the equation

$f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).$

We just need to plug in the known values and calculate the value of $$f\left( {3.5} \right):$$

$L\left( x \right) = f\left( 3 \right) + f^\prime\left( 3 \right)\left( {x - 3} \right) = 12 - 2\left( {x - 3} \right) = 18 - 2x.$

Then

$f\left( {3.5} \right) \approx 18 - 2 \cdot 3.5 = 11.$

### Example 2.

Let $$g\left( x \right)$$ be a differentiable function such that $$g\left( { - 2} \right) = 0,$$ $$g^\prime\left( { - 2} \right) = - 5.$$ Find the value of $$g\left( { - 1.8} \right)$$ using the local approximation at $$a = - 2.$$

Solution.

First, we derive the linear approximation equation for the given function:

$g\left( x \right) \approx L\left( x \right) = g\left( a \right) + g^\prime\left( a \right)\left( {x - a} \right),$

where $$a = -2.$$

Hence

$L\left( x \right) = 0 + \left( { - 5} \right)\left( {x - \left( { - 2} \right)} \right) = - 5x - 10.$

This gives the following approximate value of $$g\left( { - 1.8} \right):$$

$g\left( { - 1.8} \right) \approx - 5 \cdot \left( { - 1.8} \right) - 10 = - 1.$

### Example 3.

Find an approximate value for $$\sqrt[3]{{30}}.$$

Solution.

By the condition, $$x =30$$. We take the start point $$a = 27.$$ Then $$\Delta x = x - a = 30 - 27 = 3.$$ The derivative of the function $$f\left( x \right) = \sqrt[3]{x}$$ is given by

$f'\left( x \right) = \left( {\sqrt[3]{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}},$

and its value at the point $$a$$ is equal to

$f'\left( {a} \right) = \frac{1}{{3\sqrt[3]{{{{27}^2}}}}} = \frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.$

As a result, we get the following answer:

$f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\Delta x,\;\; \Rightarrow \sqrt[3]{{30}} \approx \sqrt[3]{{27}} + \frac{1}{{27}} \cdot 3 = 3 + \frac{1}{9} = \frac{{28}}{9} \approx 3,111.$

### Example 4.

Estimate $$\sqrt[3]{9}$$ using a linear approximation at $$a = 8.$$

Solution.

Let $$f\left( x \right) = \sqrt[3]{x}.$$ The linear approximation at the point $$a = 8$$ is given by

$f\left( x \right) \approx L\left( x \right) = f\left( 8 \right) + f^\prime\left( 8 \right)\left( {x - 8} \right).$

Find the derivative:

$f^\prime\left( x \right) = \left( {\sqrt[3]{x}} \right)^\prime = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.$

Compute the value of the derivative at $$a = 8:$$

$f^\prime\left( 8 \right) = \frac{1}{{3\sqrt[3]{{{8^2}}}}} = \frac{1}{{12}}.$

Substituting this, we get the function $$L\left( x \right)$$ in the form

$f\left( x \right) \approx L\left( x \right) = 2 + \frac{1}{{12}}\left( {x - 8} \right) = \frac{x}{{12}} + \frac{4}{3}.$

Hence

$\sqrt[3]{9} \approx L\left( 9 \right) = \frac{9}{{12}} + \frac{4}{3} = \frac{{9 + 16}}{{12}} = \frac{{25}}{{12}}.$

### Example 5.

Calculate an approximate value of $$\sqrt {50}.$$

Solution.

Consider the function $$f\left( x \right) = \sqrt x .$$ In our case, it is necessary to find the value of this function at $$x = 50.$$

We choose $$a = 49$$ and find the value of the derivative at this point:

$f\left( x \right) = \sqrt x ,\;\; \Rightarrow f'\left( x \right) = \left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},\;\; \Rightarrow f'\left( {a = 49} \right) = \frac{1}{{2\sqrt {49} }} = \frac{1}{{14}}.$

Using the formula

$f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\Delta x,$

we obtain:

$\sqrt {50} \approx \sqrt {49} + \frac{1}{{14}} \cdot \left( {50 - 49} \right) = 7 + \frac{1}{{14}} = \frac{{99}}{{14}} \approx 7,071.$

### Example 6.

Estimate $$\sqrt {3.9}$$ using a linear approximation at $$a = 4.$$

Solution.

We consider the linear approximation of $$f\left( x \right) = \sqrt x$$ at the point $$a = 4.$$

The linearization of $$f\left( x \right)$$ at $$a = 4$$ is given by

$f\left( x \right) \approx L\left( x \right) = f\left( 4 \right) + f^\prime\left( 4 \right)\left( {x - 4} \right).$

Calculate the value of the function and its derivative at this point.

$f^\prime\left( x \right) = \left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},$
$f\left( 4 \right) = \sqrt 4 = 2,$
$f^\prime\left( 4 \right) = \frac{1}{{2\sqrt 4 }} = \frac{1}{4}.$

Plug this in the equation for $$L\left( x \right):$$

$L\left( x \right) = 2 + \frac{1}{4}\left( {x - 4} \right) = 1 + \frac{x}{4}.$

$\sqrt {3.9} \approx L\left( {3.9} \right) = 1 + \frac{{3.9}}{4} = 1.975$

### Example 7.

Calculate an approximate value for $$\sqrt[4]{{0,025}}.$$

Solution.

Here it is convenient to take the value $$a = 0,0256,$$ since

$f\left( {a} \right) = \sqrt[4]{{a}} = \sqrt[4]{{0,0256}} = 0,4.$

Find the derivative of this function and its value at the point $$a:$$

$f\left( x \right) = \sqrt[4]{x},\;\; \Rightarrow f'\left( x \right) = \left( {\sqrt[4]{x}} \right)^\prime = \left( {{x^{\frac{1}{4}}}} \right)^\prime = \frac{1}{4}{x^{ - \frac{3}{4}}} = \frac{1}{{4\sqrt[4]{{{x^3}}}}},\;\; \Rightarrow f'\left( {a = 0,0256} \right) = \frac{1}{{4\sqrt[4]{{0,{{0256}^3}}}}} = \frac{1}{{4{{\left( {\sqrt[4]{{0,0256}}} \right)}^3}}} = \frac{1}{{4 \cdot 0,{4^3}}} = \frac{1}{{4 \cdot 0,064}} = \frac{1}{{0,256}} \approx 3,9063.$

Hence, we obtain the following approximate value of the function:

$f\left( x \right) \approx L\left( x \right) = f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow \sqrt[4]{{0,025}} \approx 0,4 + 3,9063 \cdot \left( {0,025 - 0,0256} \right) = 0,4 + 3,9063 \cdot \left( { - 0,0006} \right) \approx 0,3977.$

### Example 8.

Find the linearization of the function $f\left( x \right) = \sqrt[3]{{{x^2}}}$ at $$a = 27.$$

Solution.

We apply the formula

$f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right),$

where

$f\left( a \right) = f\left( {27} \right) = \sqrt[3]{{{{27}^2}}} = 9.$

Take the derivative using the power rule:

$f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime = \left( {{x^{\frac{2}{3}}}} \right)^\prime = \frac{2}{3}{x^{\frac{2}{3} - 1}} = \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{2}{{3\sqrt[3]{x}}}.$

Then

$f^\prime\left( a \right) = f^\prime\left( {27} \right) = \frac{2}{{3\sqrt[3]{{27}}}} = \frac{2}{9}.$

Substitute this in the equation for $$L\left( x \right):$$

$L\left( x \right) = 9 + \frac{2}{9}\left( {x - 27} \right) = 9 + \frac{2}{9}x - 6 = \frac{2}{9}x + 3.$

$y = \frac{2}{9}x + 3.$

### Example 9.

Calculate $${\left( {8,2} \right)^{\frac{2}{3}}}.$$

Solution.

Here, obviously, $$f\left( x \right) = {x^{\frac{2}{3}}}$$ and $$x = 8,2.$$ Let $${a = 8}.$$ Then

$f\left( {a = 8} \right) = 8^{\frac{2}{3}} = \left( {\sqrt[3]{8}} \right)^2 = 2^2 = 4.$

Find the derivative:

$f'\left( x \right) = {\left( {{x^{\frac{2}{3}}}} \right)^\prime } = \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{2}{{3\sqrt[3]{x}}},\;\; \Rightarrow f'\left( {a = 8} \right) = \frac{2}{{3\sqrt[3]{8}}} = \frac{\cancel{2}}{{3 \cdot \cancel{2}}} = \frac{1}{3}.$

The result is

$f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow {\left( {8,2} \right)^{\frac{2}{3}}} \approx 4 + \frac{1}{3} \cdot \left( {8,2 - 8} \right) = 4 + \frac{1}{3} \cdot 0,2 \approx 4,067.$

### Example 10.

Derive the approximate formula ${\left( {1 + \alpha } \right)^n} \approx 1 + n\alpha .$ Calculate the approximate value for $$\sqrt {1,02} .$$

Solution.

Consider the function $$f\left( x \right) = {x^n}.$$ When the independent variable changes by $$\Delta x$$, the increment of the function is given by

$\Delta y = {\left( {x + \Delta x} \right)^n} - {x^n}.$

If $$\Delta x$$ is a small quantity, we can approximately assume that

$\Delta y \approx dy = f'\left( x \right)\Delta x = {\left( {{x^n}} \right)^\prime }\Delta x = n{x^{n - 1}}\Delta x.$

Consequently,

${\left( {x + \Delta x} \right)^n} \approx {x^n} + n{x^{n - 1}}\Delta x.$

Suppose further that $$x = 1$$ and $$\Delta x = \alpha.$$ Then

${\left( {1 + \alpha } \right)^n} \approx {1^n} + n \cdot {1^{n - 1}} \cdot \alpha = 1 + n\alpha .$

In particular,

$\sqrt {1,02} = \sqrt {1 + 0,02} = {\left( {1 + 0,02} \right)^{\frac{1}{2}}} \approx 1 + \frac{1}{2} \cdot 0,02 = 1,01.$

See more problems on Page 2.