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bixtya [17]
2 years ago
11

If the local linear approximation of f(x) = 4x + e2x at x = 1 is used to find the approximation for f(1.1), then the % error of

this approximation is between 0% and 4% between 5% and 10% between 11% and 15% greater than 15%
Mathematics
1 answer:
castortr0y [4]2 years ago
4 0

The % error of this approximation is between 5 % and 10 %.

<h3>Procedure - Relative error of a linear approximation respect to a function</h3>

In this question we must use the function given on statement and concepts of <em>linear</em> approximation and <em>relative</em> error, which are described below:

<h3>Linear approximation</h3>

f_{e}(x + \Delta x) = f(x) +f'(x) \cdot \Delta x (1)

<h3>Relative error</h3>

e = \frac{|f_{e}(x + \Delta x)-f(x + \Delta x)|}{f(x + \Delta x)}\times 100\,\% (2)

Where:

  • f(x) - Original function evaluated at x.
  • f'(x) - First derivative of the original function evaluated at x.
  • \Delta x - Incremental change on x.
  • f_{e}(x+\Delta x) - Estimated value for x + \Delta x.
  • e - Relative error, in percentage.

If we know that f(x) = 4\cdot x + e^{2\cdot x}, f'(x) = 4 + 2\cdot e^{2\cdot x}, x = 1 and \Delta x = 0.1, then the error of the linear approximation relative to the original function is:

<h3>Original function at x = 1</h3>

f(1) = 4\cdot (1) + e^{2\cdot (1)}

f(1) \approx 11.389

<h3>First derivative of the original function at x = 1</h3>

f'(1) = 4 + 2\cdot e^{2\cdot (1)}

f'(1) \approx 11.389

<h3>Original function at x = 1.1</h3>

f(1.1) = 4\cdot (1.1) + e^{2\cdot (1.1)}

f(1.1)\approx 13.425

<h3>Linear approximation at x = 1.1</h3>

f_{e}(1.1) = 11.389 + (11.389)\cdot (0.1)

f_{e}(1.1) = 12.528

<h3>Relative error</h3>

e = \frac{|12.528 - 13.425|}{13.425}\times 100\,\%

e \approx 6.682\,\%

The % error of this approximation is between 5 % and 10 %. \blacksquare

To learn more on relative error, we kindly invite to check this verified question: brainly.com/question/13370015

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