Question

If the local linear approximation of f(x) = 2cos x + e^2x at x = 2 is used to find the approximation for f(2.1), then the % error of this approximation is?

Answer 1

Answer-

The % error of this approximation is 1.64%

Solution-

Here,

$\Rightarrow f(x) = 2\cos x + e^{2x}$

$\Rightarrow f'(x) = -2\sin x + 2e^{2x}$

And,

$\Rightarrow f(2) = 2\cos 2 + e^{4}$

$\Rightarrow f'(2) = -2\sin 2 + 2e^{4}$

Taking (2, f(2)) as a point and slope as, f'(2), the function would be,

$\Rightarrow y-y_1=m(x-x_1)$

$\Rightarrow y-(2\cos 2 + e^{4})=(-2\sin 2 + 2e^{4})(x-2)$

$\Rightarrow y=(-2\sin 2 + 2e^{4})(x-2)+(2\cos 2 + e^{4})$

The value of f(2.1) will be

$\Rightarrow y=(-2\sin 2 + 2e^{4})(2.1-2)+(2\cos 2 + e^{4})$

$\Rightarrow y=(-2\sin 2 + 2e^{4})(0.1)+(2\cos 2 + e^{4})$

$\Rightarrow y=64.5946$

According to given function, f(2.1) will be,

$\Rightarrow f(2.1) = 2\cos 2.1 + e^{2(2.1)}$

$\Rightarrow f(2.1) = 65.6766$

$\therefore \%\ error=\dfrac{65.6766-64.5946}{65.6766}=0.0164=1.64\%$