Answer-
The % error of this approximation is 1.64%
Solution-
Here,
[tex]\Rightarrow f(x) = 2\cos x + e^{2x}[/tex]
[tex]\Rightarrow f'(x) = -2\sin x + 2e^{2x}[/tex]
And,
[tex]\Rightarrow f(2) = 2\cos 2 + e^{4}[/tex]
[tex]\Rightarrow f'(2) = -2\sin 2 + 2e^{4}[/tex]
Taking (2, f(2)) as a point and slope as, f'(2), the function would be,
[tex]\Rightarrow y-y_1=m(x-x_1)[/tex]
[tex]\Rightarrow y-(2\cos 2 + e^{4})=(-2\sin 2 + 2e^{4})(x-2)[/tex]
[tex]\Rightarrow y=(-2\sin 2 + 2e^{4})(x-2)+(2\cos 2 + e^{4})[/tex]
The value of f(2.1) will be
[tex]\Rightarrow y=(-2\sin 2 + 2e^{4})(2.1-2)+(2\cos 2 + e^{4})[/tex]
[tex]\Rightarrow y=(-2\sin 2 + 2e^{4})(0.1)+(2\cos 2 + e^{4})[/tex]
[tex]\Rightarrow y=64.5946[/tex]
According to given function, f(2.1) will be,
[tex]\Rightarrow f(2.1) = 2\cos 2.1 + e^{2(2.1)}[/tex]
[tex]\Rightarrow f(2.1) = 65.6766[/tex]
[tex]\therefore \%\ error=\dfrac{65.6766-64.5946}{65.6766}=0.0164=1.64\%[/tex]
