How do I show that $-\frac{1}{e^x + 1} + 1 = \frac{e^x}{e^x + 1}$?

Question

The expression is $$-\frac{1}{e^x + 1} + 1 = \frac{e^x}{e^x + 1}$$

I would like help to get from the left side to the right side.

Answer 1

One can write the $1=\frac{e^x+1}{e^x+1}$ in $\frac{-1}{e^x+1}+1$ to obtain $$\frac{-1}{e^x+1}+\frac{e^x+1}{e^x+1}=\frac{-1+e^x+1}{e^x+1}=\frac{e^x}{e^x+1}.$$

Hope this helped!

Answer 2

$$-\frac{1}{e^x + 1} + 1 = -\frac{1}{e^x + 1} + \frac{e^x+1}{e^x+1}= \frac{-1+e^x+1}{e^x+1} = \frac{e^x}{e^x+1}{}$$

Answer 3

Simply compute the LHS.. $$-\frac{1}{e^x + 1} + 1 = \frac{-1 + e^x + 1}{e^x + 1} = \frac{e^x}{e^x + 1}$$

Answer 4

you should take lcm: see my answer

$$-\frac{1}{e^x+1}+1=-\frac{1}{e^x+1}+\frac{1}{1}=\frac{-1+1*(e^x+1)}{1*(e^x+1)}=\frac{-1+e^x+1}{e^x+1}=\frac{e^x}{e^x+1}$$