How do you solve for x?: #e^(-x)+x = 1/3?#

1 Answer

Answer:

This equation has no real solutions.

Explanation:

Let's rewrite the equation as
#e^(-x)+x-1/3=0#

We can show that the function #f(x)=e^(-x)+x-1/3# has a global minimum whose value is positive and so the former equation #f(x)=0# has no real solutions.
To find the extrema of the function we look at the boundary of the domain (in this case, the reals #RR#) and to the stationary points.

#f# is continuous on #RR# and #lim_{x to pm infty} f(x)=+infty#. This means that #f# has no global maximum and at least one global minimum.
#f'(x)=-e^[-x]+1# is null when #e^(-x)=1#, i.e. there's only one stationary point in #x=0#. This has to be the global minimum, because the extrema are always stationary points and we already guaranteed the existence of a global minimum.

#min_{x in RR} f(x) = f(0)=e^0+0-1/3=1-1/3=2/3>0#, so the function #f# is always positive and never crosses the #x#-axis.
We can conclude that #f(x)=0# doesn't have real solutions.