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Linear approximation in mathematics is an approximation of general functioning. The linear approximation function is just about finding the tangent line of the function at a specific point in a different way. Linear approximation formula is used in order to find the approximation of a function’s value at a given point.
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Key Takeaways: Linear approximation formula, linear approximation, tangent line, linear equation, function, tangent line, Derivative
Linear Approximation Formula
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The Linear Approximation formula of function f(x) is:
f(x) ≈ f(x0)+f′(x0)(x−x0)
f(x0) is the value of f(x) at x = x0.
f'(x0) is the derivative value of f(x) at x = x0.
Significance of Linear Approximation
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It often happens that in applications a model produces equations that are extremely hard or impossible to solve. But, some of the factors are more important than others. There is often a parameter, let's say x, whose values are typically small, corresponding to one of these less important factors. If you set x to 0, thus ignoring that factor completely, it simplifies the situation so much that the solution becomes easy. Therefore if you're looking for F[x], you calculate F[0].
But you don't want to ignore the factorization completely, so the next thing to try is a linear approximation. Even though you can't calculate F[x] when x≠0, you may be able to find F′(0), and thus you can use the linear approximation F[0] + xF′[0], which should produce a good approximation to F[x] when x is small.
In order to get better results in Linear approximations, quadratic and higher-order approximations must be applied.
Applications of Linear Approximation to Measurement "Error"
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Most experiments use instruments to take measurements, but the instruments used for these experiments are not perfect, and the measurements or results we get from them are not accurate or only accurate up to a certain level.
We may use the principle of a linear approximation to estimate the amount of accuracy of findings we derive from our measurements if we know the level of precision of our equipment and measurements.
Also Read: Uses of formulas
Things to Remember
- Linear approximation formula is a function that is used to approximate the value of a function at the nearest values of a fixed value.
- It is based on the equation of the tangent line of a function at a fixed point.
- Linear approximation formula is also used to estimate the amount of accuracy of findings and measurement.
- Differential is used in calculus to refer to a small change in some varying quantity.
- Derivative is used to find the rate of change in a quantity in relation to the other quantity.
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Sample Questions
Ques- Find a linear approximation to f(x)=3xe2x−10 at x=5. [2 marks]
Ans: We’ll need the derivative first as well as a couple of function evaluations.
f′(x)=3e2x−10+6xe2x−10 f(5)=15 f′(5)=33
write down the linear approximation.
L(x)=15+33(x−5)=33x−150
Ques- Find a linear approximation to h(t)=t4−6t3+3t−7 at t=−3. [3 marks]
Ans: We’ll need the derivative first as well as a couple of function evaluations.
h′(t)=4t3−18t2+3 h(−3)=227 h′(−3)=−267
write down the linear approximation.
L(t)=227−267(t+3)=−267t−574
Ques- Calculate the linear approximation of the function f(x) = x2 as the value of x tends to 2 ? [3 marks]
Ans:
Given,
f(x) = x2 x0 = 2
f(x0) = 22 = 4
f ‘(x) = 2x
f'(x0) = 2(2) = 4
Linear approximation formula is,
f(x)≈f(x0)+f′(x0)(x−x0) f(x) ≈ 4+ 4( 2- 2)
f(x) ≈ 4 +4(0)
f(x) =4
Ques- Find the approximate value of ln 1.1 and ln 0.9 using linear approximation. [5 marks]
Ans: We will use the function f(x) = ln x, and the value a = 1. Then
f'(x) = 1/x.
f'(a) = f'(1) = 1.
Furthermore, f(a) = ln 1 = 0.
Therefore, the linear approximation of f at x = 1 is
y = f(1) + f'(1) (x - 1)
y = 0 + 1 (x - 1)
y = x - 1
The linear approximation gives us the following values:
f(1.1) = ln 1.1 can be approximated by
y = 1.1 - 1 = 0.1.
f(0.9) can be approximated by
y = 0.9 - 1 = -0.1.
Now, a calculator shows us that ln 1.1 is approximately 0.09531 and ln 0.9 approximately -1054.
These are fairly accurate approximations.
Ques- Find the equation of linear approximation of the function f(x) = cos x at x = π/2. [5 marks]
Ans: The given function is, f(x) = cos x.
We have to find the linear approximation of f(x) at a = π/2.
So f(a) = cos π/2 = 0.
The derivative of f(x) is,
f ' (x) = - sin x
f ' (a) = f ' (π/2) = - sin π/2 = -1.
The linear approximation formula of f(x) is,
L(x) = f(a) + f '(a) (x - a)
L(x) = 0 + (-1) (x - π/2)
L(x) = -x + π/2
The equation of linear approximation is, L(x) = -x + π/2.
Ques- Use the linear approximation formula to find the approximate value of 27.05.Round your answer to 4 decimals. [3 marks]
Ans: Let us assume that f(x) = √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) = √27 = 3.
The derivative of f(x) is,
f ' (x) = d/dx (√x) = d/dx (x1/3) = 1/3 . x -âÂ\(\Box\)Â\(\Box\)
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,
27.05.≈ L(27.05) = (1/27)(27.05) + 2 = 3.0019
27.05 ≈ 3.0019.
Ques- How do you use linear Approximation to find the value of (1.01)10? [3 marks]
Ans: We let f(x)=x10. Then f'(x)=10x9
Now use the linear approximation formula, which is y= f(a) + f'(a) (x−a for a function f(x) approximated at a point close to x=a.
Our point will obviously be a=1.
Also, f(1)=1, f'(1)=10 and x−a=1.01−1=0.01
Plugging this into our formula, we have:
y= 1+10 (0.01) = 1.1
Ques- Find the linear approximation of the function g(x) = 51 + 3x at x = 0. [5 marks]
Ans: Solve for the value of the given linear function for g(0) and g'(0). Solve for the value of the function and its derivative at a = 0.
g(a) = 51 + 3(0)
g(0) = 51 + 0
g(0) = 51
g’(a) = 51 + 3x
g’(0) = 0 + 3
g’(0) = 3
In getting the approximate value of g(x) = 51 + x, apply the approximate formula or the tangent line equation.
L(x) = f(a) + f’(a) (x - a)
L(x) = g(a) + g’(a) (x - a)
L(x) = g(0) + g’(0) (x - 0)
L(x) = 51 + 1(0 – 0)
L(x) = 51
Therefore, the approximate value of the given function g(x) = 51 + 3x at x = 0 is 51.
Ques- Use linear approximation, i.e. the tangent line, to approximate (1/0.252) as follows Let f(x)= 1/x and find the equation of the tangent line in slope-intercept form to f(x) at a "nice" point near 0.252 [3 marks]
Ans:
f(.25) = 4
f'(x) = -1/x2
f'(.25) = - 16
Equation of tangent (in pt-slope form): y - 4 = - 16(x - .25). in slope-int form : y = -16x + 8
y(.252) = 3.968
∴ f(.252) ≈ 3.968
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