Question

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?

Answer 1

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`