1. Problem 7: Find the domain and range of the relation $R = \{(x,y) \mid y \leq x-1 \text{ and } y \geq 2x-1\}$. - The domain is the set of all $x$ values for which there exists $y$ such that $2x -1 \leq y \leq x -1$. - Since $y$ lies between the two inequalities, for the set to be non-empty, $2x -1 \leq x -1$ must hold. - Simplify: $2x -1 \leq x -1 \Rightarrow 2x -1 \leq x -1$ implies $2x \leq x$ which simplifies to $x \leq 0$. - Therefore, the domain is $\boxed{(-\infty, 0]}$. - For range, fix any $x \leq 0$; then $y$ lies between $2x -1$ and $x -1$. - The smallest possible $y$ is $2x -1$, the largest is $x -1$. - For $x=0$, $y$ ranges from $-1$ to $-1$, so $y = -1$. - To find overall range, consider the smallest lower bound and largest upper bound over all $x \leq 0$. - As $x \to -\infty$, $2x -1 \to -\infty$ and $x -1 \to -\infty$, so $y$ unbounded below. - The maximum upper bound occurs at $x=0$, $y=-1$. - Thus the range is $(-\infty, -1]$. --- 2. Problem 8: Sketch the graph of $f(x) = 3^x \operatorname{sgn}(x)$ and find: i) Domain and range. - The sign function $\operatorname{sgn}(x)$ is defined as: $$\operatorname{sgn}(x) = \begin{cases} -1 & x < 0 \\ 0 & x = 0 \\ 1 & x > 0 \end{cases}$$ - So: - For $x<0$, $f(x) = 3^x \times (-1) = -3^x$. - For $x=0$, $f(0)=3^0 \times 0 = 1 \times 0 = 0$. - For $x>0$, $f(x)=3^x \times 1 = 3^x$. - The domain of $f$ is all real numbers, since $3^x$ is defined for all real $x$, and $\operatorname{sgn}(x)$ is defined everywhere. - The range consists of: - Negative values approaching $0$ from below for $x<0$ since $-3^x$ ranges from $-1$ (at $x=0^-$) to $-\infty$ (as $x \to -\infty$). - The point $0$ at $x=0$. - Positive values from $1$ to $+\infty$ for $x > 0$. - Writing range more completely: - For $x<0$, $f(x) \in (-\infty, -1)$. - At $x=0$, $f(0) = 0$. - For $x>0$, $f(x) \in (1, +\infty)$. - Notice the function is discontinuous at $0$ since limits from left and right differ: $$\lim_{x \to 0^-} f(x) = -1, \quad f(0) = 0, \quad \lim_{x \to 0^+} f(x) = 1$$ - Hence range is: $$(-\infty, -1) \cup \{0\} \cup (1, \infty)$$ --- 3. Problem 9: Identify parity of given functions. Parity means: - Even if $f(-x) = f(x)$, - Odd if $f(-x) = -f(x)$, - Neither if neither condition holds. a) $f(x) = x^2 \operatorname{sgn}(x)$ - Analyze sign change: - $f(-x) = (-x)^2 \operatorname{sgn}(-x) = x^2 (-\operatorname{sgn}(x)) = -x^2 \operatorname{sgn}(x) = -f(x)$. - Thus $f$ is odd. b) $f(x) = |x| + 2$ - $f(-x) = |-x| + 2 = |x| + 2 = f(x)$. - So $f$ is even. c) $f(x) = x^3 + 2x$ - Check $f(-x) = (-x)^3 + 2(-x) = -x^3 - 2x = - (x^3 + 2x) = -f(x)$. - So $f$ is odd. --- 4. Problem 10: Let $f(x) = \frac{2}{3} x^{-1/2}$. a) Domain: - $x^{-1/2} = \frac{1}{\sqrt{x}}$. - So $x$ must be > 0 to have real values. - Therefore, domain is $(0, \infty)$. b) Range: - $f(x) = \frac{2}{3} \frac{1}{\sqrt{x}}$. - As $x \to 0^+$, $\sqrt{x} \to 0^+$, so $f(x) \to +\infty$. - As $x \to \infty$, $\sqrt{x} \to \infty$, so $f(x) \to 0^+$. - Since function is positive and continuous on $(0,\infty)$, range is $(0, \infty)$. --- 5. Problem 11: Find domain and range of inverse of relation $R=\{(x,y): y \leq x^2 -1 \text{ and } y \leq 4\}$. - First, rewrite $R$: $$y \leq x^2 -1 \quad \text{and} \quad y \leq 4$$ implies $y \leq \min(x^2 -1, 4)$. - Domain of $R$ in $x$ is $\mathbb{R}$ (no restriction). - Range is all $y$ satisfying $y \leq \min(x^2 -1,4)$. - Since $x^2 -1$ is ≤ 4 when $x^2 -1 \leq 4 \Rightarrow x^2 \leq 5$. - For $|x| \leq \sqrt{5}$, $y \leq x^2 -1 \leq 4$. - For $|x| > \sqrt{5}$, $x^2 -1 > 4$, so $y \leq 4$. - So range is $(-\infty, 4]$ because $y \leq$ min is always less than or equal to 4. - Inverse relation swaps roles of $x$ and $y$: $$R^{-1} = \{(y,x) \mid y \leq x^2 -1, y \leq 4\} \Rightarrow (x,y) \in R^{-1} \iff x \leq y^2 -1, x \leq 4.$$ But it is symmetric in notation, so we treat $y$ as input variable. - Thus domain of $R^{-1}$ is all $y$ such that $x \leq y^2 -1$ and $x \leq 4$ holds. - Equivalently, for the inverse, domain is $y$ with no restriction (since $x$ varies), and range is $x \leq \min(y^2 -1, 4)$. - Domain of $R^{-1}$ is $\mathbb{R}$. - Range is $(-\infty,4]$. --- 6. Problem 12: Sketch the graph of floor function $f(x) = |x-2|$. - The floor function $[f(x)] = \lfloor |x - 2| \rfloor$ maps $f(x)$ to the greatest integer less than or equal to $|x-2|$. - Domain: all real numbers. - Range: all integers $\geq 0$ (because absolute value is always non-negative). - The graph looks like steps starting at $x=2$ where $f(x)=0$, increasing by 1 at points where $|x-2|$ passes an integer. - For example: - At $x=2$, value is $0$, - For $1 \leq x < 2$, floor values step at 0,1, etc., mirrored on the other side. - The graph consists of horizontal segments on each interval corresponding to $|x-2|$ lying in $[n, n+1)$ for integers $n$.