How to prove that #e^x-x# is decreasing for x<0, without using derivatives?

How to prove that #e^x-x# is decreasing for #x<0#, without using derivatives?

1 Answer
Aug 9, 2017

There are two parts to consider, #e^x# and #-x#.

With #e^x#, the greater the negative nunber, the smaller the value. #e^(-a)=1/e^a#, as #x# approaches #-oo#, #e^x# approaches #0#, and therefore has little effect on #-x#, #-x>e^x#.

Therefore, the smaller the value, the less impact on #-x#, and the more it resembles #-x#, which has a constant decreasing gradient.

#e^x-x,# # x>0# will be increasing due to #e^x>##-x#