The % error of this approximation is between 5 % and 10 %.
Procedure - Relative error of a linear approximation respect to a function
In this question we must use the function given on statement and concepts of linear approximation and relative error, which are described below:
Linear approximation
[tex]f_{e}(x + \Delta x) = f(x) +f'(x) \cdot \Delta x[/tex] (1)
Relative error
[tex]e = \frac{|f_{e}(x + \Delta x)-f(x + \Delta x)|}{f(x + \Delta x)}\times 100\,\%[/tex] (2)
Where:
- [tex]f(x)[/tex] - Original function evaluated at [tex]x[/tex].
- [tex]f'(x)[/tex] - First derivative of the original function evaluated at [tex]x[/tex].
- [tex]\Delta x[/tex] - Incremental change on [tex]x[/tex].
- [tex]f_{e}(x+\Delta x)[/tex] - Estimated value for [tex]x + \Delta x[/tex].
- [tex]e[/tex] - Relative error, in percentage.
If we know that [tex]f(x) = 4\cdot x + e^{2\cdot x}[/tex], [tex]f'(x) = 4 + 2\cdot e^{2\cdot x}[/tex], [tex]x = 1[/tex] and [tex]\Delta x = 0.1[/tex], then the error of the linear approximation relative to the original function is:
Original function at x = 1
[tex]f(1) = 4\cdot (1) + e^{2\cdot (1)}[/tex]
[tex]f(1) \approx 11.389[/tex]
First derivative of the original function at x = 1
[tex]f'(1) = 4 + 2\cdot e^{2\cdot (1)}[/tex]
[tex]f'(1) \approx 11.389[/tex]
Original function at x = 1.1
[tex]f(1.1) = 4\cdot (1.1) + e^{2\cdot (1.1)}[/tex]
[tex]f(1.1)\approx 13.425[/tex]
Linear approximation at x = 1.1
[tex]f_{e}(1.1) = 11.389 + (11.389)\cdot (0.1)[/tex]
[tex]f_{e}(1.1) = 12.528[/tex]
Relative error
[tex]e = \frac{|12.528 - 13.425|}{13.425}\times 100\,\%[/tex]
[tex]e \approx 6.682\,\%[/tex]
The % error of this approximation is between 5 % and 10 %. [tex]\blacksquare[/tex]
To learn more on relative error, we kindly invite to check this verified question: https://brainly.com/question/13370015
