1. Statement of the problem: State the given function $f(x)=3e^{-4x}+5$ for $x\in\mathbb{R}$. 2. Objective: (i) Find the range of $f$. (ii) Find $f^{-1}$ and state its domain. 3. Formula and rules: For inverses of functions of form $y=a e^{bx}+c$ we solve for $x$ by isolating the exponential and applying the natural logarithm $\ln$. Important rules: - Exponential $e^{t}>0$ for all real $t$, so expressions inside $\ln$ must be positive. - Domain of $f^{-1}$ equals range of $f$, and range of $f^{-1}$ equals domain of $f$. 4. Range of $f$ (part i): Consider limits: as $x\to\infty$, $e^{-4x}\to0$ so $f(x)\to5$. As $x\to-\infty$, $e^{-4x}\to\infty$ so $f(x)\to\infty$. Because $3e^{-4x}>0$ for all $x$, values of $f(x)$ are always greater than 5, and can be arbitrarily large. Therefore the range is $(5,\infty)$. 5. Find the inverse (part ii): Start with $y=3e^{-4x}+5$. Subtract 5 from both sides: $y-5=3e^{-4x}$. Divide both sides by 3: $$\frac{y-5}{\cancel{3}}=\frac{3e^{-4x}}{\cancel{3}}$$ which simplifies to $\frac{y-5}{3}=e^{-4x}$. Apply natural logarithm: $\ln\left(\frac{y-5}{3}\right)=-4x$. Divide both sides by -4: $$\frac{-4x}{\cancel{-4}}=\frac{\ln\left(\frac{y-5}{3}\right)}{\cancel{-4}}$$ so $x=-\frac{1}{4}\ln\left(\frac{y-5}{3}\right)$. Replace $y$ with $x$ to write the inverse function: $f^{-1}(x)=-\frac{1}{4}\ln\left(\frac{x-5}{3}\right)$. 6. Domain of $f^{-1}$: The argument of $\ln$ must be positive, so $\frac{x-5}{3}>0$, which gives $x-5>0$ and thus $x>5$. Therefore the domain of $f^{-1}$ is $(5,\infty)$. 7. Final answers: (i) Range of $f$: $(5,\infty)$. (ii) $f^{-1}(x)=-\frac{1}{4}\ln\left(\frac{x-5}{3}\right)$ with domain $(5,\infty)$. Notes about the graph: The horizontal asymptote is $y=5$ and the curve decays from high values on the left towards 5 on the right, matching the description.