1. **State the problem:** We are given the function $$f(x) = 3e^{-4x} + 5$$ for all real numbers $$x \in \mathbb{R}$$. We need to: (i) Find the range of $$f$$. (ii) Find the inverse function $$f^{-1}$$ and state its domain. 2. **Recall important properties:** - The exponential function $$e^t$$ is always positive for any real $$t$$. - For $$f(x) = a e^{bx} + c$$, if $$a > 0$$ and $$b < 0$$, the function is an exponential decay shifted vertically by $$c$$. - The range of $$f$$ is determined by the horizontal asymptote and the behavior of the exponential term. 3. **Find the range of $$f$$:** - Since $$e^{-4x} > 0$$ for all $$x$$, the smallest value of $$3e^{-4x}$$ is greater than 0. - As $$x \to \infty$$, $$e^{-4x} \to 0$$, so $$f(x) \to 5$$ from above. - As $$x \to -\infty$$, $$e^{-4x} \to \infty$$, so $$f(x) \to \infty$$. Therefore, the range is: $$\boxed{(5, \infty)}$$ 4. **Find the inverse function $$f^{-1}$$:** Start with: $$y = 3e^{-4x} + 5$$ Isolate the exponential term: $$y - 5 = 3e^{-4x}$$ Divide both sides by 3: $$\frac{y - 5}{3} = e^{-4x}$$ Take the natural logarithm of both sides: $$\ln\left(\frac{y - 5}{3}\right) = -4x$$ Divide both sides by -4: $$x = -\frac{1}{4} \ln\left(\frac{y - 5}{3}\right)$$ So the inverse function is: $$f^{-1}(y) = -\frac{1}{4} \ln\left(\frac{y - 5}{3}\right)$$ 5. **State the domain of $$f^{-1}$$:** - The argument of the logarithm must be positive: $$\frac{y - 5}{3} > 0 \implies y - 5 > 0 \implies y > 5$$ Thus, the domain of $$f^{-1}$$ is: $$\boxed{(5, \infty)}$$ --- **Final answers:** - Range of $$f$$: $$(5, \infty)$$ - Inverse function: $$f^{-1}(y) = -\frac{1}{4} \ln\left(\frac{y - 5}{3}\right)$$ - Domain of $$f^{-1}$$: $$(5, \infty)$$