Answer is: option3
\( 0.453 \)Solution:
The Lagrange error is
\[ R_n(x)=\frac{f^{(n+1)}(c)(x-a)^{\,n+1}}{(n+1)!} \]
For \( f(x)=e^{x/2} \):
\[ f'(x)=\frac{1}{2}e^{x/2}, \qquad f''(x)=\frac{1}{4}e^{x/2}, \qquad f'''(x)=\frac{1}{8}e^{x/2} \]
Since \( f'''(x) \) is increasing, positive, and continuous on \( [0,2] \), the maximum occurs at \( x=2 \):
\[ f'''(2)=\frac{1}{8}e^{2/2}=\frac{e}{8} \]
When \( n=2,\; a=0,\; x=2 \):
\[ R_2(x)= \frac{\frac{e}{8}(2-0)^3}{3!} = \frac{e}{6} \approx 0.453 \]
