## University Calculus: Early Transcendentals (3rd Edition)

(a) $f^{-1}=\log_2\frac{x}{100-x}$ (b) $f^{-1}=\log_{1.1}\frac{x}{50-x}$
(a) $$y=f(x)=\frac{100}{1+2^{-x}}$$ 1. Solve for $x$ in terms of $y$: $$y=\frac{100}{1+2^{-x}}$$ $$1+2^{-x}=\frac{100}{y}$$ $$2^{-x}=\frac{100}{y}-1=\frac{100-y}{y}$$ Here we take $\log_2$ of both sides: $$-x=\log_2\frac{100-y}{y}$$ $$x=-\log_2\frac{100-y}{y}=\log_2\Big(\frac{100-y}{y}\Big)^{-1}=\log_2\frac{y}{100-y}$$ 2. Switch $x$ and $y$: $$y=\log_2\frac{x}{100-x}$$ So $$f^{-1}=\log_2\frac{x}{100-x}$$ 3. Test: $$f(f^{-1})=\frac{100}{1+2^{-\log_2\frac{x}{100-x}}}=\frac{100}{1+2^{\log_2\frac{100-x}{x}}}=\frac{100}{1+\frac{100-x}{x}}$$ $$f(f^{-1})=\frac{100}{\frac{x+100-x}{x}}=\frac{100x}{100}=x$$ $$f^{-1}(f)=\log_2\frac{\frac{100}{1+2^{-x}}}{100 - \frac{100}{1+2^{-x}}}=\log_2\frac{\frac{100}{1+2^{-x}}}{\frac{100+100\times2^{-x}-100}{1+2^{-x}}}=\log_2\frac{100}{100\times2^{-x}}$$ $$f^{-1}(f^)=\log_2\frac{1}{2^{-x}}=\log_22^x=x$$ Hence, $f(f^{-1})=f^{-1}(f)=x$ (b) $$y=f(x)=\frac{50}{1+1.1^{-x}}$$ 1. Solve for $x$ in terms of $y$: $$y=\frac{50}{1+1.1^{-x}}$$ $$1+1.1^{-x}=\frac{50}{y}$$ $$1.1^{-x}=\frac{50}{y}-1=\frac{50-y}{y}$$ Here we take $\log_{1.1}$ of both sides: $$-x=\log_{1.1}\frac{50-y}{y}$$ $$x=-\log_{1.1}\frac{50-y}{y}=\log_{1.1}\Big(\frac{50-y}{y}\Big)^{-1}=\log_{1.1}\frac{y}{50-y}$$ 2. Switch $x$ and $y$: $$y=\log_{1.1}\frac{x}{50-x}$$ So $$f^{-1}=\log_{1.1}\frac{x}{50-x}$$ 3. Test: $$f(f^{-1})=\frac{50}{1+1.1^{-\log_{1.1}\frac{x}{50-x}}}=\frac{50}{1+1.1^{\log_{1.1}\frac{50-x}{x}}}=\frac{50}{1+\frac{50-x}{x}}$$ $$f(f^{-1})=\frac{50}{\frac{x+50-x}{x}}=\frac{50x}{50}=x$$ $$f^{-1}(f)=\log_{1.1}\frac{\frac{50}{1+1.1^{-x}}}{50 - \frac{50}{1+1.1^{-x}}}=\log_{1.1}\frac{\frac{50}{1+1.1^{-x}}}{\frac{50+50\times1.1^{-x}-50}{1+1.1^{-x}}}=\log_{1.1}\frac{50}{50\times1.1^{-x}}$$ $$f^{-1}(f^)=\log_{1.1}\frac{1}{1.1^{-x}}=\log_{1.1}1.1^x=x$$ Hence, $f(f^{-1})=f^{-1}(f)=x$