Thread: [SOLVED] Formula for inverse function

1. [SOLVED] Formula for inverse function

Let f(x)=20/(1+3e^-x). Find a formula for f^-1(x).

So I tried to do it this way:

y=f(x)=20/(1+3e^-x).

x=20/(1+3e^-y)

20/x=1+3e^-y

log ((20/x)-1)/log(3e)

2. Hello, CalcQueen!

You were okay up to the last step(s) (which you didn't show).

Let $\displaystyle f(x)\,=\,\frac{20}{1+3e^{-x}}$ . Find $\displaystyle f^{-1}(x)$

So I tried to do it this way:

. . $\displaystyle y\:=\:\frac{20}{1+3e^{-x}}$

. . $\displaystyle x\:=\:\frac{20}{1+3e^{-y}}$

. . $\displaystyle 1+3e^{-y} \:=\:\frac{20}{x}$ . . . . good!

. . $\displaystyle \frac{\log\left (\frac{20}{x}-1\right)}{\log(3e)}$ . ?

You had: .$\displaystyle 1 + 3e^{-y} \:=\:\frac{20}{x}$

Then: .$\displaystyle 3e^{-y} \:=\:\frac{20}{x} - 1 \quad\Rightarrow\quad 3e^{-y}\:=\:\frac{20-x}{x} \quad\Rightarrow\quad e^{-y} \:=\:\frac{20-x}{3x}$

Take logs: .$\displaystyle \ln\left(e^{-y}\right) \:=\:\ln\left(\frac{20-x}{3x}\right) \quad\Rightarrow\quad (-y)\ln(e) \:=\:\ln\left(\frac{20-x}{3x}\right)$

Since $\displaystyle \ln(e) = 1$, we have: .$\displaystyle -y \:=\:\ln\left(\frac{20-x}{3x}\right)$

Finally: .$\displaystyle y \:=\:-\ln\left(\frac{20-x}{3x}\right) \quad\Rightarrow\quad y \:=\:\ln\left(\frac{20-x}{3x}\right)^{-1} \quad\Rightarrow\quad\boxed{ y \:=\:\ln\left(\frac{3x}{20-x}\right)}$