The evaluation of the function [tex]f(x)=e^x[/tex] at a=0 and x=5 by the taylor series will have a value 65.37.
We know the fromula of taylor series is given as:
[tex]f(x)=f(a)+\dfrac{f'((a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3.................[/tex]
Now for the function [tex]f(x)=e^x[/tex] the taylor's series will become.
[tex]f(x)=e^a+e^ax+\dfrac{e^ax^2}{2!}+\dfrac{e^ax^3}{3!}+\dfrac{e^ax^4}{4!}[/tex]
[tex]f(x)=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}[/tex]
So at x=5 the values will become
[tex]f(x)=1+5+12.5+20.83+26.04=65.37[/tex]
hence the evaluation of the function [tex]f(x)=e^x[/tex] at a=0 and x=5 by the taylor series will have a value 65.37.
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