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

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=(65.6766-64.5946)/(65.6766)=0.0164=1.64\%`