Inverse Functions — Question 2

Question 2

Let $f(x) = \ln(x - 1)$, where $x > 1$.

• Find the inverse function $f^{-1}(x)$.

• State the domain and range of $f(x)$ and $f^{-1}(x)$.

• Verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Question 2 - Solution

(a) Find the inverse of $f(x) = \ln(x - 1)$:

Let $y = \ln(x - 1)$

Exponentiate both sides: $$e^y = x - 1 \Rightarrow x = e^y + 1$$

Now swap $x$ and $y$ to get the inverse: $$f^{-1}(x) = e^x + 1$$

Answer: $\boxed{f^{-1}(x) = e^x + 1}$

(b) Domain and Range:

For $f(x) = \ln(x - 1)$:

Domain: $x > 1 \Rightarrow \boxed{(1, \infty)}$ , Range of $\ln(x - 1)$ is all real numbers: $\boxed{(-\infty, \infty)}$

So for the inverse:

Domain of $f^{-1}(x)$: $\boxed{(-\infty, \infty)}$ , Range of $f^{-1}(x) = e^x + 1$: since $e^x > 0$, we get $f^{-1}(x) > 1 \Rightarrow \boxed{(1, \infty)}$

Summary: $$\begin{aligned} \text{Domain of } f(x) &= (1, \infty) \\ \text{Range of } f(x) &= (-\infty, \infty) \\ \text{Domain of } f^{-1}(x) &= (-\infty, \infty) \\ \text{Range of } f^{-1}(x) &= (1, \infty) \end{aligned}$$

(c) Verify compositions:

$$f(f^{-1}(x)) = \ln((e^x + 1) - 1) = \ln(e^x) = x$$

$$f^{-1}(f(x)) = e^{\ln(x - 1)} + 1 = x - 1 + 1 = x$$

Both identities verified.