Inverse Functions — Question 9

Question 9 - Solution

(a) Find the inverse:

Let $y = \ln(x - 2)$

Solve for $x$:

$$x - 2 = e^y \Rightarrow x = e^y + 2$$

Now switch $x$ and $y$:

$$f^{-1}(x) = e^x + 2$$

Answer: $\boxed{f^{-1}(x) = e^x + 2}$

(b) Domain and Range:

For $f(x) = \ln(x - 2)$: - Domain: $x > 2 \Rightarrow \boxed{(2, \infty)}$ - Range: $\ln(x - 2) \to -\infty$ as $x \to 2^+$, and $\to \infty$ as $x \to \infty$ $$\text{Range: } \boxed{(-\infty, \infty)}$$

For $f^{-1}(x) = e^x + 2$: - Domain: $\boxed{(-\infty, \infty)}$ - Range: $e^x + 2 > 2 \Rightarrow \boxed{(2, \infty)}$

(c) Verify compositions:

$$f(f^{-1}(x)) = f(e^x + 2) = \ln((e^x + 2) - 2) = \ln(e^x) = x$$

$$f^{-1}(f(x)) = f^{-1}(\ln(x - 2)) = e^{\ln(x - 2)} + 2 = x - 2 + 2 = x$$

Verified: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$