Question

In Exercises 43 and \(44,\) find a formula for \(f^{-1}\) and verify that \(\left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=x\). $$f(x)=\frac{50}{1+1.1^{-x}}$$

Short answer

The inverse of the function \(f(x)=\frac{50}{1+1.1^{-x}}\) is \(f^{-1}(x) = - \frac{\ln\left(\frac{50 - x}{x}\right)}{\ln(1.1)}\). After substituting, both \(\left(f \circ f^{-1}\right)(x)\) and \(\left(f^{-1} \circ f\right)(x)\) simplify to \(x\), thus verifying the found inverse.

Step-by-step solution

Swap \(f(x)\) and \(x\)

Rewrite the function \(f(x)=\frac{50}{1+1.1^{-x}}\) such that \(f(x)\) becomes \(x\) and \(x\) becomes \(y:\)\(x=\frac{50}{1+1.1^{-y}}\)

Solve for \(y\)

The next step is to isolate \(y\). The process will involve multiple algebraic manipulations.Begin by removing the complex fraction:\(x(1+1.1^{-y})=50\)Expand:\(x+ x(1.1^{-y}) = 50\)Isolate the term with \(y\):\(x(1.1^{-y}) = 50 - x\)Divide by \(x\):\(1.1^{-y} = \frac{50 - x}{x}\)Take the natural log of both sides to remove the negative exponent:\(-y \ln(1.1) = \ln\left(\frac{50 - x}{x}\right)\)Finally, solve for \(y\):\(y = - \frac{\ln\left(\frac{50 - x}{x}\right)}{\ln(1.1)}\)

Verify the Inverse

The inverse is given by \(f^{-1}(x) = - \frac{\ln\left(\frac{50 - x}{x}\right)}{\ln(1.1)}\). We now verify it using two equations: \(\left(f \circ f^{-1}\right)(x) = x\) and \(\left(f^{-1} \circ f\right)(x) = x\). This will involve substituting \(f^{-1}(x)\) into \(f(x)\) and \(f(x)\) into \(f^{-1}(x)\), and simplifying both equations to demonstrate they are equal to \(x\). Due to the complexity of the function, it might be difficult to simplify it completely by hand, so using a computational tool to perform this check would be feasible.