Determine if $f(x) = \frac{100}{1+2^{-x}}$ is one-to-one

Question

I currently have $$f(x) = \frac{100}{1+2^{-x}}$$ and I'm trying to determine if it is one-to-one. I have looked at the other posts however I'm not sure as to how to deal with the $-x$.

Answer 1

If $f(x)=\dfrac{100}{1+2^{-x}}$ then $2^{-x}=\dfrac{100}{f(x)}-1$ so $x=-\log_{2}\dfrac{100}{f(x)}=\log_2 f(x)-\log_2 100$.

$f$ is injective, as if $f(x_1)=f(x_2)$ then $x_1=-\log_{2}\dfrac{100}{f(x_1)}=-\log_{2}\dfrac{100}{f(x_1)}=x_2$, because $\log_2$ is injective. Whether $f$ is surjective or not depends on the Range you choose for $f$.

Answer 2

It is because $y=2^{-x}$ is. An exponential function to any positive base is. You could take the derivative, and check that it's strictly decreasing.

Also, $y=\dfrac{100}{1+x}$ is one-to-one. So we have the composition of two one-to-one functions.