Find the approximate value of e^ 1.005 . Given that e=2.7183..

Question

Find the approximate value of $e^{1.005}$. Given that $e=2.7183$..

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Answer

Let $f(x)=e^x$
Differentiate w. r. $t . x$.
$f^{\prime}(x)=e^x$
Let $a=1, h=0.005$
For $x=a=1$, from (I) we get
$f(a)=f(1)=e^1=2.7183$
For $x=a=1$, from (II) we get
$
f^{\prime}(a)=f^{\prime}(1)=e^1=2.7183
$
We have, $f(a+h) \doteqdot f(a)+h f^{\prime}(a)$
$
\begin{gathered}
f(1+0.005) \doteqdot f(1)+(0.005) \cdot f^{\prime}(1) \\
f(1.005) \doteqdot 2.7183+(0.005)(2.7183) \ldots \\
\ldots[\text { From (III) and (IV)] } \\
f(1.005) \doteqdot 2.7183+0.0135915 \\
\doteqdot 2.7318915 \\
f(1.005)=e^{1.005} \doteqdot 2.73189
\end{gathered}
$

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