1. **Determine whether the following statements are functions of $x$ or not.** (i). $y = x^2 + 1$ for each $x \in \mathbb{R}$. - A function assigns exactly one output for each input. - Here, for each real number $x$, $y$ is uniquely defined as $x^2 + 1$. - So, this is a function. (ii). $y = |x - 1|$ for each $x \in \mathbb{R}$. - The absolute value function also assigns exactly one output for each input. - For every real $x$, $y$ is uniquely $|x - 1|$. - So, this is a function. (iii). $y^2 = 20x$ for each $x \geq 0$. - Here, $y$ is defined implicitly. - For a given $x \geq 0$, $y = \pm \sqrt{20x}$, so two possible $y$ values. - Since one input $x$ can give two outputs $y$, this is not a function. 2. **Briefly sketch the graphs of the following functions:** (i). $y = \sqrt{x}$ for each $x > 0$. - The graph starts at $(0,0)$ and increases gradually to the right. - It is only defined for $x \geq 0$. (ii). $y = x^3 + 1$ for each $x \in \mathbb{R}$. - This is a cubic curve shifted up by 1. - Passes through $(-1,0)$, $(0,1)$, $(1,2)$. (iii). $y = |3x - 5|$ for each $x \in \mathbb{R}$. - This is a V-shaped graph with vertex at $x = \frac{5}{3}$. - Opens upwards. 3. **Find the domains of the following functions:** (i). $f(x) = \sqrt{x - 5} + x^2 + 1$ - The square root requires $x - 5 \geq 0$. - So, domain is $[5, \infty)$. (ii). $f(x) = \frac{x - 3}{2x - 5}$ - Denominator cannot be zero: $2x - 5 \neq 0 \Rightarrow x \neq \frac{5}{2}$. - Domain is $\mathbb{R} \setminus \left\{\frac{5}{2}\right\}$. (iii). $f(x) = \frac{x}{\sqrt{x - 5}} + \frac{5}{3 - x}$ - $\sqrt{x - 5}$ requires $x - 5 > 0 \Rightarrow x > 5$. - Denominator $3 - x \neq 0 \Rightarrow x \neq 3$. - Since $x > 5$, $x \neq 3$ is automatically satisfied. - Domain is $(5, \infty)$. 4. **Is $f(x) = x^3 + 5$ a bijection?** - A function is bijection if it is both injective (one-to-one) and surjective (onto). - $f$ is strictly increasing cubic function, so injective. - For any real $y$, there exists $x = \sqrt[3]{y - 5}$ such that $f(x) = y$, so surjective. - Hence, $f$ is bijection. 5. **Does $f(x) = x^2 + 3$ have an inverse?** - $f: \mathbb{R} \to \mathbb{R}$ is not injective because $f(-x) = f(x)$. - So, $f^{-1}$ does not exist on $\mathbb{R}$. For $g: [0, \infty) \to [0, \infty)$ defined by $g(x) = x^2 + 3$: - Domain is restricted to $x \geq 0$ making $g$ injective. - Inverse exists and is $g^{-1}(y) = \sqrt{y - 3}$ for $y \geq 3$. 6. **Is $f(x) = |x - 5|$ an injection and/or surjection?** - Injection: No, because $f(4) = f(6) = 1$ but $4 \neq 6$. - Surjection: No, codomain is $\mathbb{R}$ but $f(x) \geq 0$, so negative values are not attained. 7. **For $f: [0,81) \to [-3,6]$ defined by $f(x) = \sqrt{x - 3}$:** (i). Injection? - $f$ is not defined for $x < 3$ (since $x-3$ under root). - Domain should be $[3,81)$. - $f$ is increasing on $[3,81)$, so injective. (ii). Surjection? - $f(3) = 0$, $f(81) = \sqrt{78} \approx 8.83$. - Codomain is $[-3,6]$, but $f(x) \geq 0$, so not surjective onto $[-3,6]$. (iii). For $g: [0,25] \to [-2,3]$ defined by $g(x) = \sqrt{x} - 2$: - $g(0) = -2$, $g(25) = 5 - 2 = 3$. - $g$ is increasing, so surjective onto $[-2,3]$. Final answers: (i) function, (ii) function, (iii) not function. Domains: (i) $[5,\infty)$, (ii) $\mathbb{R} \setminus \{5/2\}$, (iii) $(5,\infty)$. $f(x)=x^3+5$ is bijection. $f(x)=x^2+3$ no inverse on $\mathbb{R}$, inverse exists on $[0,\infty)$ as $g^{-1}(y)=\sqrt{y-3}$. $f(x)=|x-5|$ not injection, not surjection. $f(x)=\sqrt{x-3}$ injective but not surjective onto $[-3,6]$. $g(x)=\sqrt{x}-2$ surjective onto $[-2,3]$.