Question

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%

Answer 1

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

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

Relative error

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

Original function at x = 1

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

$f(1) \approx 11.389$

First derivative of the original function at x = 1

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

$f'(1) \approx 11.389$

Original function at x = 1.1

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

$f(1.1)\approx 13.425$

Linear approximation at x = 1.1

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

$f_{e}(1.1) = 12.528$

Relative error

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

$e \approx 6.682\,\%$

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

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