Question

Hello everyone. Solve for x.
$\rm \leadsto {e}^{ - x} + x - 2 = 0$​

Answer 1

Step-by-step explanation:

I think it won't be easy to solve this using normal algebra so I used graphical approach.

Here is your answer.

Answer 2

$\large\textsf{Given equation is : \rm{e^{-x}+x-2} }$

$\bullet\:\:\textrm{Lets draw the graph of }\rm{f(x)=e^{-x}+x-2}$

$\dashrightarrow\textrm{Kindly Refer to Attachment for Graph}\dashleftarrow$

$\textrm{Solving the equation \rm{e^{-x}+x-2=0} means to find the \rm{x} - intercepts of \rm{e^{-x}+x-2=0} }$$\textrm{The graph of the function \rm{e^{-x}+x-2=0} crosses the \rm{x} axis at points (-1.146,0) \& (1.841,0)}$

$\textrm{Hence, the \rm{x} -intercepts of function \rm{e^{-x}+x-2=0} are \rm{x=-1.146} and \rm{x=1.841} }$$\textrm{So, the solution of the equation \rm{e^{-x}+x-2=0} is \mathrm{x}=-1.146 or \mathrm{x}=1.841 }$