1. Problem: Determine if each function is invertible and find the inverse if it exists. 2. Function 1: $f: \mathbb{R} \to \mathbb{R}^+, f(x) = x^2$ - Rule: A function is invertible if it is one-to-one (injective). - $f(x) = x^2$ is not one-to-one on $\mathbb{R}$ because $f(-a) = f(a)$. - Therefore, $f$ is not invertible on $\mathbb{R}$. 3. Function 2: $f: \mathbb{R}^+ \to \mathbb{R}^+, f(x) = \frac{5}{x}$ - Domain and codomain are positive real numbers. - $f$ is one-to-one and onto. - To find inverse, solve $y = \frac{5}{x}$ for $x$: $$ y = \frac{5}{x} \implies x = \frac{5}{y} $$ - So inverse function is $f^{-1}(y) = \frac{5}{y}$. 4. Function 3: $f: \mathbb{R}^+ \to [2, \infty), f(x) = x^2 + 2$ - $f$ is one-to-one on $\mathbb{R}^+$ because $x^2$ is increasing for $x > 0$. - To find inverse, solve $y = x^2 + 2$ for $x$: $$ y - 2 = x^2 \implies x = \sqrt{y - 2} $$ - So inverse function is $f^{-1}(y) = \sqrt{y - 2}$. 5. Function 4: $f: \mathbb{R} \setminus \{3\} \to \mathbb{R} \setminus \{1\}, f(x) = \frac{x-1}{x-3}$ - Check if $f$ is one-to-one: Suppose $f(a) = f(b)$: $$ \frac{a-1}{a-3} = \frac{b-1}{b-3} $$ Cross-multiplied: $$ (a-1)(b-3) = (b-1)(a-3) $$ Expanding and simplifying leads to $a = b$. - So $f$ is one-to-one. - To find inverse, solve $y = \frac{x-1}{x-3}$ for $x$: $$ y(x-3) = x - 1 $$ $$ yx - 3y = x - 1 $$ $$ yx - x = 3y - 1 $$ $$ x(y - 1) = 3y - 1 $$ $$ x = \frac{3y - 1}{y - 1} $$ - So inverse function is $f^{-1}(y) = \frac{3y - 1}{y - 1}$. 6. Function 5: $f: [0, \pi] \to [3, -3], f(x) = 3 \cos(x)$ - Note codomain is $[3, -3]$ which is decreasing interval; usually codomain should be $[-3, 3]$. - $\cos(x)$ is decreasing on $[0, \pi]$ from 1 to -1, so $3 \cos(x)$ decreases from 3 to -3. - $f$ is one-to-one on $[0, \pi]$. - To find inverse, solve $y = 3 \cos(x)$ for $x$: $$ \cos(x) = \frac{y}{3} $$ $$ x = \arccos\left(\frac{y}{3}\right) $$ - So inverse function is $f^{-1}(y) = \arccos\left(\frac{y}{3}\right)$. Final answers: - Function 1: Not invertible. - Function 2: Inverse $f^{-1}(y) = \frac{5}{y}$. - Function 3: Inverse $f^{-1}(y) = \sqrt{y - 2}$. - Function 4: Inverse $f^{-1}(y) = \frac{3y - 1}{y - 1}$. - Function 5: Inverse $f^{-1}(y) = \arccos\left(\frac{y}{3}\right)$.