1. Problem: Determine if the function $f: \mathbb{R} \to \mathbb{R}^+$ defined by $f(x) = x^2$ is invertible. 2. To check invertibility, a function must be one-to-one (injective) and onto (surjective). 3. For $f(x) = x^2$ over $\mathbb{R}$, it is not one-to-one because $f(-a) = f(a)$ for any $a \in \mathbb{R}$. 4. Therefore, $f$ is not invertible on $\mathbb{R}$. --- 1. Problem: Determine if $f: \mathbb{R}^+ \to \mathbb{R}^+$ defined by $f(x) = \frac{5}{x}$ is invertible and find its inverse. 2. Since $f(x)$ is defined on positive reals and $f(x) = \frac{5}{x}$ is strictly decreasing and continuous, it is one-to-one and onto $\mathbb{R}^+$. 3. To find the inverse, set $y = \frac{5}{x}$. 4. Solve for $x$: $x = \frac{5}{y}$. 5. Thus, the inverse function is $f^{-1}(y) = \frac{5}{y}$. --- 1. Problem: Determine if $f: \mathbb{R}^+ \to [2, \infty)$ defined by $f(x) = x^2 + 2$ is invertible and find its inverse. 2. Since $x^2$ is strictly increasing on $\mathbb{R}^+$ and adding 2 shifts the range to $[2, \infty)$, $f$ is one-to-one and onto. 3. To find the inverse, set $y = x^2 + 2$. 4. Solve for $x$: $x = \sqrt{y - 2}$. 5. So, $f^{-1}(y) = \sqrt{y - 2}$. --- 1. Problem: Determine if $f: \mathbb{R} \setminus \{3\} \to \mathbb{R} \setminus \{1\}$ defined by $f(x) = \frac{x-1}{x-3}$ is invertible and find its inverse. 2. The function is rational and one-to-one on its domain. 3. Set $y = \frac{x-1}{x-3}$. 4. Solve for $x$: $$y(x-3) = x - 1$$ $$yx - 3y = x - 1$$ $$yx - x = 3y - 1$$ $$x(y - 1) = 3y - 1$$ $$x = \frac{3y - 1}{y - 1}$$ 5. The inverse function is $f^{-1}(y) = \frac{3y - 1}{y - 1}$. --- 1. Problem: Determine if $f: [0, \pi] \to [3, -3]$ defined by $f(x) = 3 \cos(x)$ is invertible. 2. The function $3 \cos(x)$ on $[0, \pi]$ decreases from 3 to -3, so it is one-to-one and onto. 3. To find the inverse, set $y = 3 \cos(x)$. 4. Solve for $x$: $$\cos(x) = \frac{y}{3}$$ $$x = \arccos\left(\frac{y}{3}\right)$$ 5. So, $f^{-1}(y) = \arccos\left(\frac{y}{3}\right)$. Final answers: - $f(x) = x^2$ on $\mathbb{R}$ is not invertible. - $f(x) = \frac{5}{x}$ on $\mathbb{R}^+$ is invertible with inverse $f^{-1}(y) = \frac{5}{y}$. - $f(x) = x^2 + 2$ on $\mathbb{R}^+$ is invertible with inverse $f^{-1}(y) = \sqrt{y - 2}$. - $f(x) = \frac{x-1}{x-3}$ on $\mathbb{R} \setminus \{3\}$ is invertible with inverse $f^{-1}(y) = \frac{3y - 1}{y - 1}$. - $f(x) = 3 \cos(x)$ on $[0, \pi]$ is invertible with inverse $f^{-1}(y) = \arccos\left(\frac{y}{3}\right)$.