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 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 \(\sqrt[\large 3\normalsize]{{30}}.\)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 \(\sqrt[\large 4\normalsize]{{0,025}}.\)Example 8
Find the linearization of the function \(f\left( x \right) = \sqrt[3]{{{x^2}}}\) at \(a = 27.\)Example 9
Calculate \({\left( {8,2} \right)^{\large\frac{2}{3}\normalsize}}.\)Example 10
Derive the approximate formula \({\left( {1 + \alpha } \right)^n} \approx 1 + n\alpha .\) Calculate the approximate value for \(\sqrt {1,02} .\)Example 11
Derive the approximate formula \[{\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}}\;\;}\kern-0.3pt{\left( {a \gt 0} \right).}\] Using this identity calculate approximately the value of \(\sqrt {150} .\)Example 12
Derive the approximate formula \[{\sqrt[\large n\normalsize]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n – 1}}}}\;\;}\kern-0.3pt{\left( {a \gt 0} \right).}\] Using this formula, calculate \(\sqrt[\large 8\normalsize]{{250}}.\)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 \(f\left( x \right) = \large{\frac{3}{{{x^2}}}}\normalsize.\) 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 \(f\left( x \right) = \large{\frac{{{x^2} + 1}}{{x – 2}}}\normalsize\) 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 \(x = 180.\)- Find the coordinates \(\left( {a,f\left( a \right)} \right)\) of the point \(P.\)
- Write the linearization equation \(L\left( x \right).\)
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 \(\sqrt[\large 3\normalsize]{{30}}.\)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 \(f\left( x \right) = \sqrt[\large 3\normalsize]{x}\) is given by
\[
{f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } }
= {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } }
= {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} }
= {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}
\]
and its value at the point \(a\) is equal to
\[
{f’\left( {a} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} }
= {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}
\]
As a result, we get the following answer:
\[
{f\left( x \right) \approx f\left( {a} \right) + f’\left( {a} \right)\Delta x,\;\;}\Rightarrow
{\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 }
= {3 + \frac{1}{9} }
= {\frac{{28}}{9} \approx 3,111.}
\]
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 \(L\left( x \right):\)
\[{L\left( x \right) = 2 + \frac{1}{4}\left( {x – 4} \right) }={ 1 + \frac{x}{4}.}\]
So, the answer is
\[{\sqrt {3.9} \approx L\left( {3.9} \right) }={ 1 + \frac{{3.9}}{4} }={ 1.975}\]
Example 7.
Calculate an approximate value for \(\sqrt[\large 4\normalsize]{{0,025}}.\)Solution.
Here it is convenient to take the value \(a = 0,0256,\) since
\[
{f\left( {a} \right) = \sqrt[\large 4\normalsize]{{a}} }
= {\sqrt[\large 4\normalsize]{{0,0256}} = 0,4.}
\]
Find the derivative of this function and its value at the point \(a:\)
\[
{f\left( x \right) = \sqrt[4]{x},\;\;}\Rightarrow
{f’\left( x \right) = {\left( {\sqrt[\large 4\normalsize]{x}} \right)^\prime } }
= {{\left( {{x^{\large\frac{1}{4}\normalsize}}} \right)^\prime } }
= {\frac{1}{4}{x^{ – \large\frac{3}{4}\normalsize}} }
= {\frac{1}{{4\sqrt[\large 4\normalsize]{{{x^3}}}}},\;\;}\Rightarrow
{f’\left( {a = 0,0256} \right) }
= {\frac{1}{{4\sqrt[\large 4\normalsize]{{0,{{0256}^3}}}}} }
= {\frac{1}{{4{{\left( {\sqrt[\large 4\normalsize]{{0,0256}}} \right)}^3}}} }
= {\frac{1}{{4 \cdot 0,{4^3}}} }
= {\frac{1}{{4 \cdot 0,064}} }
= {\frac{1}{{0,256}} \approx 3,9063.}
\]
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 \(L\left( x \right):\)
\[{L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}\]
Answer:
\[y = \frac{2}{9}x + 3.\]