1. Problem (a): Find the inverse function $f^{-1}$ of the function $f(x) = -3 + \sqrt{3 + 2x}$ and determine the domain and range of $f^{-1}$. 2. To find the inverse, set $y = f(x) = -3 + \sqrt{3 + 2x}$. 3. Solve for $x$ in terms of $y$: $$y = -3 + \sqrt{3 + 2x} \implies y + 3 = \sqrt{3 + 2x}$$ 4. Square both sides to eliminate the square root: $$ (y + 3)^2 = 3 + 2x $$ 5. Isolate $x$: $$ 2x = (y + 3)^2 - 3 \implies x = \frac{(y + 3)^2 - 3}{2} $$ 6. The inverse function is therefore: $$ f^{-1}(y) = \frac{(y + 3)^2 - 3}{2} $$ 7. Now find the domain and range of $f^{-1}$. Since $f^{-1}$ swaps the domain and range of $f$, first find the domain and range of $f$. 8. The domain of $f$ is all $x$ values for which the expression under the square root is non-negative: $$3 + 2x \geq 0 \implies x \geq -\frac{3}{2}$$ 9. For $x \geq -\frac{3}{2}$, compute the range of $f(x)$: $$ f(x) = -3 + \sqrt{3 + 2x} \geq -3 + 0 = -3 $$ As $x$ grows large, $f(x)$ increases without bound, so the range is: $$ \mathrm{Range}(f) = [-3, \infty) $$ 10. Therefore, the domain of $f^{-1}$ is $[-3, \infty)$ and the range of $f^{-1}$ is $[-\frac{3}{2}, \infty)$. --- 11. Problem (b): Consider $f$ satisfying $[f(x)]^5 + f(x) = \ln x$ for $x > 0$. Show that $f^{-1}$ exists and evaluate $[f(1)]^5 - [f^{-1}(1)]^5$. 12. To show $f^{-1}$ exists, we need to prove $f$ is one-to-one and onto its range. Define $g(t) = t^5 + t$. 13. Note $g'(t) = 5t^4 + 1 > 0$ for all real $t$, so $g(t)$ is strictly increasing and hence one-to-one. 14. The equation $g(f(x)) = \ln x$ implies $f(x) = g^{-1}(\ln x)$, so $f$ is the composition of an increasing function $g^{-1}$ and an increasing function $\ln x$ (for $x>0$), hence $f$ is strictly increasing and injective. 15. Since $f$ is strictly increasing and defined on $(0, \infty)$, its inverse $f^{-1}$ exists. 16. Evaluate the expression: $$ [f(1)]^5 - [f^{-1}(1)]^5 $$ 17. Given $x=1$, we have: $$ [f(1)]^5 + f(1) = \ln 1 = 0 $$ Let $a = f(1)$. Then: $$ a^5 + a = 0 \implies a(a^4 + 1) = 0 $$ 18. Since $a^4 + 1 > 0$, it follows $a=0$. Hence: $$ f(1) = 0 $$ 19. Next, find $f^{-1}(1)$, i.e., find $x$ such that: $$ f(x) = 1 $$ 20. Using the defining relation: $$ [f(x)]^5 + f(x) = \ln x $$ Substitute $f(x) = 1$: $$1^5 + 1 = \ln x \implies 2 = \ln x \implies x = e^{2}$$ 21. Therefore: $$ f^{-1}(1) = e^{2} $$ 22. Evaluate the expression: $$ [f(1)]^{5} - [f^{-1}(1)]^5 = 0^5 - (e^{2})^{5} = - e^{10} $$ Final answers: - (a) $f^{-1}(y) = \frac{(y+3)^2 -3}{2}$ with domain $[-3, \infty)$ and range $[-\frac{3}{2}, \infty)$ - (b) Inverse $f^{-1}$ exists and $$ [f(1)]^{5} - [f^{-1}(1)]^{5} = - e^{10} $$