# Linear Approximation

## Solved Problems

Click or tap a problem to see the solution.

### Example 11

Derive the approximate formula $\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}},$ where a > 0. Using this identity calculate approximately the value of $$\sqrt {150} .$$

### Example 12

Derive the approximate formula $\sqrt[n]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n - 1}}}},$ where a > 0. Using this formula, calculate $$\sqrt[8]{{250}}.$$

### Example 13

Find an approximate value for $$\cos 46^\circ.$$

### Example 14

Find the linearization of the function $f\left( x \right) = {x^2} + 2\cos x$ at $$a = 0.$$

### Example 15

Find an approximate value for $$\sin 179^\circ.$$

### Example 16

Find the linearization of the natural logarithm $f\left( x \right) = \ln x$ at $$x = 1.$$

### Example 17

Find an approximate value of $$\ln 20.$$

### Example 18

Calculate $${e^{0,1}}.$$

### Example 19

Find an approximate value for $$\arccos 0,51.$$

### Example 20

Find an approximate value for $$\arctan 0,95.$$

### Example 21

Given the function $f\left( x \right) = \frac{3}{{{x^2}}}.$ Use linear approximation at $$a = 2$$ to estimate the value of $$f\left( {1.99} \right).$$

### Example 22

Find the approximate value of the function $f\left( x \right) = \sqrt {{x^2} + 3x}$ at $$x = 1,02.$$

### Example 23

Find the approximate value of the function $f\left( x \right) = \sqrt {5x - 1}$ at $$x = 1,99.$$

### Example 24

Find the linear approximation to $f\left( x \right) = \frac{{{x^2} + 1}}{{x - 2}}$ near $$a = 3.$$

### Example 25

The function $f\left( x \right) = - {x^2} + 100x$ is linearized at a point $$P\left( {a,f\left( a \right)} \right).$$ The linear approximation line $$L\left( x \right)$$ passes through $$P$$ and intersects the $$x-$$axis at $$x = 180.$$

1. Find the coordinates $$\left( {a,f\left( a \right)} \right)$$ of the point $$P.$$
2. Write the linearization equation $$L\left( x \right).$$

### Example 11.

Derive the approximate formula $\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}},$ where $${a \gt 0}.$$ Using this identity calculate approximately the value of $$\sqrt {150} .$$

Solution.

Consider the function $$y = \sqrt x .$$ If the independent variables changes by $$\Delta x$$, the increment of the function is written as

$\Delta y = \sqrt {x + \Delta x} - \sqrt x .$

This increment for small $$\Delta x$$ can be approximated by the differential so that

$\Delta y = \sqrt {x + \Delta x} - \sqrt x \approx dy = f'\left( x \right)\Delta x = {\left( {\sqrt x } \right)^\prime }\Delta x = \frac{1}{{2\sqrt x }}\Delta x.$

Thus,

$\sqrt {x + \Delta x} \approx \sqrt x + \frac{1}{{2\sqrt x }}\Delta x.$

We denote $$x = {a^2}$$, $$\Delta x = h.$$ Then we obtain the following approximate equation:

$\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}}.$

Using this formula we can estimate the value of $$\sqrt {150}:$$

$\sqrt {150} = \sqrt {144 + 6} = \sqrt {{{12}^2} + 6} \approx 12 + \frac{1}{4} = 12,25.$

The exact value (up to $$3$$ digits after the decimal point) is equal to $$12,247$$. As it can be seen, the relative error when using the approximate formula is

$\frac{{12,250 - 12,247}}{{12,247}} = \frac{{0,003}}{{12,247}} = 2,4 \cdot {10^{ - 4}} \lt 0,03\% .$

### Example 12.

Derive the approximate formula $\sqrt[n]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n - 1}}}},$ where $${a \gt 0}.$$ Using this formula, calculate $$\sqrt[8]{{250}}.$$

Solution.

Let $$y = \sqrt[n]{x}.$$ If the variable $$x$$ changes by $$\Delta x,$$ the increment of the function has the form:

$\Delta y = \sqrt[n]{{x + \Delta x}} - \sqrt[n]{x}.$

Considering $$\Delta x$$ as a small quantity, we can replace the increment of the function $$\Delta y$$ by its differential:

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

Then

$\sqrt[n]{{x + \Delta x}} \approx \sqrt[n]{x} + \frac{1}{n}{x^{\frac{{1 - n}}{n}}}\Delta x.$

Denoting $$x = {a^n}$$, $$\Delta x = h,$$ we obtain the following relationship:

$\sqrt[n]{{{a^n} + h}} \approx a + \frac{1}{n}{\left( {{a^n}} \right)^{\frac{{1 - n}}{n}}}h = a + \frac{h}{{n{a^{n - 1}}}}.$

Using this formula, we find:

$\sqrt[8]{{250}} = \sqrt[8]{{256 - 6}} = \sqrt[8]{{{2^8} - 6}} \approx 2 + \frac{{\left( { - 6} \right)}}{{8 \cdot {2^7}}} = 2 - \frac{6}{{1024}} \approx 1,994.$

### Example 13.

Find an approximate value for $$\cos 46^\circ.$$

Solution.

We choose $$a = 45^\circ.$$ The derivative of cosine at this point is equal to

$f\left( x \right) = \cos x,\;\; \Rightarrow f'\left( x \right) = - \sin x,\;\; \Rightarrow f'\left( {a = 45^\circ} \right) = - \sin 45^\circ = - \frac{{\sqrt 2 }}{2}.$

Express the increment of the independent variable $$\Delta x$$ in radians:

$\Delta x = 46^\circ - 45^\circ = 1^\circ = \frac{{2\pi }}{{360}} = \frac{\pi }{{180}}\;\text{radians}.$

Using the approximate formula for small $$\Delta x$$

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

we get:

$\cos 46^\circ \approx \cos 45^\circ + \left( { - \frac{{\sqrt 2 }}{2}} \right) \cdot \frac{\pi }{{180}} = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} \cdot \frac{\pi }{{180}} = \frac{{\sqrt 2 }}{2}\left( {1 - \frac{\pi }{{180}}} \right) \approx 0,7071 \cdot \left( {1 - 0,0175} \right) = 0,6948.$

### Example 14.

Find the linearization of the function $f\left( x \right) = {x^2} + 2\cos x$ at $$a = 0.$$

Solution.

Use the linear approximation

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

Take the derivative:

$f^\prime\left( x \right) = \left( {{x^2} + \cos x} \right)^\prime = 2x - 2\sin x.$

Then

$f\left( a \right) = f\left( 0 \right) = {0^2} + 2\cos 0 = 2,$
$f^\prime\left( a \right) = f^\prime\left( 0 \right) = 2 \cdot 0 - 2\sin 0 = 0.$

Substitute this into $$L\left( x \right):$$

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

Thus, the linear approximation of the given function is the horizontal line

$L\left( x \right) = 2.$

### Example 15.

Find an approximate value for $$\sin 179^\circ.$$

Solution.

Let $$x = 179^\circ$$, $$a = 180^\circ.$$ Hence, $$\Delta x = x - a = 179^\circ - 180^\circ = - 1$$ $$= - {\frac{\pi }{{180}}}$$ radians. Calculate the value of the function and its derivative at the point $$a:$$

$f\left( {a} \right) = \sin 180^\circ = 0,\;\;f'\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x,\;\;f'\left( {a} \right) = \cos 180^\circ = - 1.$

Replacing the increment of the function by its differential, we obtain:

$f\left( x \right) \approx f\left( {a} \right) + dy = f\left( {a} \right) + f'\left( {a} \right)\Delta x,\;\; \Rightarrow \sin 179^\circ = 0 - 1 \cdot \left( { - \frac{\pi }{{180}}} \right) = \frac{\pi }{{180}} \approx 0,0175.$

### Example 16.

Find the linearization of the natural logarithm $f\left( x \right) = \ln x$ at $$x = 1.$$

Solution.

We need to calculate $$f\left( a \right)$$ and $$f^\prime\left( a \right)$$ where $$a = 1.$$

$f\left( a \right) = f\left( 1 \right) = \ln 1 = {0},$
$f^\prime\left( x \right) = \left( {\ln x} \right)^\prime = \frac{1}{x},$
$f^\prime\left( a \right) = f^\prime\left( 1 \right) = \frac{1}{1} = {1}.$

Write the linear approximation function $$L\left( x \right):$$

$L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right) = 0 + 1\left( {x - 1} \right) = x - 1.$

So the linearization of the natural logarithm at $$x = 1$$ is

${\ln x \approx x - 1.}$

### Example 17.

Find an approximate value of $$\ln 20.$$

Solution.

Consider the natural logarithmic function $$y = \ln x.$$ Given that

$\ln {e^3} = 3\ln e = 3,$

it is convenient to take the point $${a}$$ such that

$a = {e^3} \approx 20,086.$

Calculate the derivative and its value at the point $${a}:$$

$f'\left( x \right) = \left( {\ln x} \right)^\prime = \frac{1}{x},\;\; \Rightarrow f'\left( {a} \right) \approx \frac{1}{{20,086}} \approx 0,0498.$

Hence, the approximate value of $$\ln 20$$ is equal

$\ln 20 \approx \ln {e^3} + 0,0498 \cdot \left( {20 - 20,086} \right) = 3 - 0,0043 = 2,9957.$

### Example 18.

Calculate $${e^{0,1}}.$$

Solution.

Let $$f\left( x \right) = {e^x}.$$ By setting $$a = 0,$$ we have:

$f\left( {a} \right) = {e^0} = 1,\;\;f'\left( x \right) = \left( {{e^x}} \right)^\prime = {e^x},\;\;f'\left( {a = 0} \right) = {e^0} = 1.$

To calculate $${e^{0,1}}$$ we use the approximate formula

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

Then

${e^{0,1}} \approx 1 + 1 \cdot \left( {0,1 - 0} \right) = 1,1.$

### Example 19.

Find an approximate value for $$\arccos 0,51.$$

Solution.

Suppose that $$f\left( x \right) = \arccos x$$ and $$a = 0,5.$$ Replacing the increment of the function $$\Delta y$$ by its differential, we can compute the approximate value of $$\arccos 0,51:$$

$f\left( x \right) \approx f\left( {a} \right) + dy = f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),$
$f'\left( x \right) = {\left( {\arccos x} \right)^\prime } = - \frac{1}{{\sqrt {1 - {x^2}} }},\;\; \Rightarrow f'\left( {a = 0,5} \right) = - \frac{1}{{\sqrt {1 - 0,{5^2}} }} = - \frac{1}{{\sqrt {0,75} }} \approx - 1,1547,\;\; \Rightarrow \arccos 0,51 \approx \arccos 0,5 + \left( { - 1,1547} \right) \cdot \left( {0,51 - 0,5} \right) = \frac{\pi }{3} - 0,0115 \approx 1,035\;\text{radians} \approx 59,34^\circ.$

### Example 20.

Find an approximate value for $$\arctan 0,95.$$

Solution.

Let $$f\left( x \right) = \arctan x$$, $$a = 1.$$ Determine the value of the derivative at $$a:$$

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

For the approximate calculation we use the formula

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

Consequently,

$\arctan 0,95 \approx \arctan 1 + \frac{1}{2} \cdot \left( {0,95 - 1} \right) = \frac{\pi }{4} + \frac{1}{2} \cdot \left( { - 0,05} \right) \approx 0,7604\;\text{radians} \approx 43,57^\circ.$

### Example 21.

Given the function $f\left( x \right) = \frac{3}{{{x^2}}}.$ Use linear approximation at $$a = 2$$ to estimate the value of $$f\left( {1.99} \right).$$

Solution.

We apply the linear approximation formula

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

where $$a= 2.$$

Take the derivative:

$f^\prime\left( x \right) = \left( {\frac{3}{{{x^2}}}} \right)^\prime = \left( {3{x^{ - 2}}} \right)^\prime = - 6{x^{ - 3}} = - \frac{6}{{{x^3}}}.$

The function and the derivative have the following values at $$a = 2:$$

$f\left( a \right) = f\left( 2 \right) = \frac{3}{{{2^2}}} = \frac{3}{4},$
$f^\prime\left( a \right) = f^\prime\left( 2 \right) = - \frac{6}{{{2^3}}} = - \frac{3}{4}.$

Plugging $$a,$$ $$f\left( a \right),$$ and $$f^\prime\left( a \right),$$ we obtain:

$L\left( x \right) = \frac{3}{4} + \left( { - \frac{3}{4}} \right)\left( {x - 2} \right) = - \frac{3}{4}x + \frac{9}{4} = \frac{3}{4}\left( {3 - x} \right).$

Then the approximate value of $$f\left( {1.99} \right)$$ is equal to

$f\left( {1.99} \right) \approx L\left( {1.99} \right) = \frac{3}{4}\left( {3 - 1.99} \right) = 0.7575$

### Example 22.

Find the approximate value of the function $f\left( x \right) = \sqrt {{x^2} + 3x}$ at $$x = 1,02.$$

Solution.

Choose the point $$a = 1.$$ Then

$f\left( {a} \right) = \sqrt {{1^2} + 3 \cdot 1} = 2.$

Find the value of the derivative of the given function at $$a:$$

$f'\left( x \right) = \left( {\sqrt {{x^2} + 3x} } \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 3x}} \cdot \left( {{x^2} + 3x} \right)^\prime } = \frac{{2x + 3}}{{2\sqrt {{x^2} + 3x} }},\;\; \Rightarrow f'\left( {a = 1} \right) = \frac{{2 \cdot 1 + 3}}{{2\sqrt {{1^2} + 3 \cdot 1} }} = \frac{5}{4} = 1,25.$

Hence, the approximate value of the function at $$x = 1,02$$ is equal to

$f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow \sqrt {1,02} \approx 2 + 1,25 \cdot \left( {1,02 - 1} \right) = 2,025.$

### Example 23.

Find the approximate value of the function $f\left( x \right) = \sqrt {5x - 1}$ at $$x = 1,99.$$

Solution.

Let $$a = 2$$. Hence,

$f\left( {a = 2} \right) = \sqrt {5 \cdot 2 - 1} = \sqrt 9 = 3.$

The derivative at $$a = 2$$ has the following value:

$f'\left( x \right) = {\left( {\sqrt {5x - 1} } \right)^\prime } = \frac{1}{{2\sqrt {5x - 1} }} \cdot {\left( {5x - 1} \right)^\prime } = \frac{5}{{2\sqrt {5x - 1} }},\;\; \Rightarrow f'\left( {a = 2} \right) = \frac{5}{{2\sqrt {5 \cdot 2 - 1} }} = \frac{5}{6} \approx 0,833.$

Estimating the approximate value of the function at $$x = 1,99$$, we have

$f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\; \Rightarrow {\sqrt {1,99}} \approx {3 + 0,833 \cdot \left( {1,99 - 2} \right) } = 3 - 0,0083 \approx 2,992.$

### Example 24.

Find the linear approximation to $f\left( x \right) = \frac{{{x^2} + 1}}{{x - 2}}$ near $$a = 3.$$

Solution.

Let's linearize this function at $$a = 3.$$ First, take the derivative using the quotient rule and the chain rule.

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

Calculate $$f\left( a \right)$$ and $$f^\prime\left( a \right):$$

$f\left( 3 \right) = \frac{{{3^2} + 1}}{{3 - 2}} = 10,$
$f^\prime\left( 3 \right) = \frac{{{3^2} - 4 \cdot 3 - 1}}{{{{\left( {3 - 2} \right)}^2}}} = - 4.$

Now we can write the linear approximation equation:

$f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right) = 10 - 4\left( {x - 3} \right) = - 4x + 22.$

Thus, the linearization of the given function at $$a = 3$$ is

$y = - 4x + 22.$

### Example 25.

The function $f\left( x \right) = - {x^2} + 100x$ is linearized at a point $$P\left( {a,f\left( a \right)} \right).$$ The linear approximation line $$L\left( x \right)$$ passes through $$P$$ and intersects the $$x-$$axis at $$x = 180.$$

1. Find the coordinates $$\left( {a,f\left( a \right)} \right)$$ of the point $$P.$$
2. Write the linearization equation $$L\left( x \right).$$

Solution.

$$1.$$ The linear approximation $$L\left( x \right)$$ is given by the equation

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

Take the derivative:

$f^\prime\left( x \right) = \left( { - {x^2} + 100x} \right)^\prime = - 2x + 100.$

At the point $$P,$$ the equation for $$L\left( x \right)$$ becomes

$L\left( x \right) = - {a^2} + 100a + \left( { - 2a + 100} \right)\left( {x - a} \right) = - {a^2} + \cancel{100a} - 2ax + 100x + 2{a^2} - \cancel{100a} = {a^2} + \left( {100 - 2a} \right)x.$

We find the value of $$a$$ from the condition $$L\left( x \right) = 0$$ at $$x = 180.$$ This yields:

${a^2} + \left( {100 - 2a} \right) \cdot 180 = 0,\;\; \Rightarrow {a^2} - 360a + 18000 = 0.$

Solve the quadratic equation:

$D = {360^2} - 4 \cdot 18000 = 57600.$
$a = \frac{{360 \pm \sqrt {57600} }}{2} = \frac{{360 \pm 240}}{2} = 60,300.$

We see that only one root $$a = 60$$ belongs to the interval $$\left( {0,100} \right),$$ so the point $$P$$ has the coordinates:

$a = 60,\; f\left( a \right) = - {60^2} + 100 \cdot 60 = 2400.$

$$2.$$ The equation of the linear approximation $$L\left( x \right)$$ derived in Section $$A$$ has the form

$L\left( x \right) = {a^2} + \left( {100 - 2a} \right)x.$

Substituting $$a = 60,$$ we get

$L\left( x \right) = {60^2} + \left( {100 - 2 \cdot 60} \right)x = 3600 - 20x.$