Solve #e^x+x+1=0# ?

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Answer 1 · 2018-02-10T20:03:11

See below.

You can calculate this root with the help of the so called Lambert function #W(cdot)#

We have

#e^x+x+1=0#

making #y = x+1# we have

#e^(y-1) + y = 0# or

#e^(-y)(-y) = e^-1#

now using the fact

#Y = X e^X rArr X = W(Y)#

we have

#-y = W(e^-1)#

then

# -(x+1) = W(e^-1) rArr x = -1-W(e^-1) = -1.278464542761074#