1. Problem statement: Given $f(x)=3e^{-4x}+5$ for $x\in\mathbb{R}$.\n2. What is asked: (i) State the range of $f$.\n(ii) Find $f^{-1}$ and state its domain.\n3. Formula and important rule: For $a>0$ and real $b$, the function $g(x)=a e^{b x}+c$ has range $(c,\infty)$ because $e^{t}\in(0,\infty)$ for every real $t$.\n4. Determine the range of $f$.\nThe exponential part satisfies $e^{-4x}\in(0,\infty)$ for all $x\in\mathbb{R}$.\nMultiplying by 3 gives $3e^{-4x}\in(0,\infty)$.\nAdding 5 shifts the interval to $(5,\infty)$.\nTherefore the range of $f$ is $(5,\infty)$.\n5. Find the inverse function $f^{-1}$.\nLet $y=f(x)$ so $y=3e^{-4x}+5$.\nSubtract 5 from both sides to isolate the exponential: $$y-5=3e^{-4x}$$\nDivide both sides by 3: $$\frac{3e^{-4x}}{3}=\frac{y-5}{3}$$\nShow cancellation when dividing by 3: $$\frac{\cancel{3}e^{-4x}}{\cancel{3}}=\frac{y-5}{3}$$\nThus $$e^{-4x}=\frac{y-5}{3}$$\nTake natural logarithm of both sides: $$-4x=\ln\left(\frac{y-5}{3}\right)$$\nDivide both sides by -4: $$\frac{-4x}{-4}=\frac{\ln\left(\frac{y-5}{3}\right)}{-4}$$\nShow cancellation when dividing by -4: $$\frac{\cancel{-4}x}{\cancel{-4}}=\frac{\ln\left(\frac{y-5}{3}\right)}{-4}$$\nSo $$x=-\frac{1}{4}\ln\left(\frac{y-5}{3}\right)$$\nReplace $y$ by $x$ to write the inverse function in standard form: $$f^{-1}(x)=-\frac{1}{4}\ln\left(\frac{x-5}{3}\right)$$\n6. Domain of the inverse: The domain of $f^{-1}$ equals the range of $f$, which is $(5,\infty)$.\n7. Final answers:\n(i) Range of $f$ is $(5,\infty)$.\n(ii) $f^{-1}(x)=-\dfrac{1}{4}\ln\left(\dfrac{x-5}{3}\right)$ with domain $(5,\infty)$.\n