1. Problem: Find the solution set for \(|x-5| \leq 4\). Step 1: Recall that \(|x-a| \leq b\) means \(a-b \leq x \leq a+b\). Step 2: Substitute \(a=5\) and \(b=4\). Step 3: So, \(5-4 \leq x \leq 5+4\) which gives \(1 \leq x \leq 9\). Answer: The solution set is \([1,9]\). 2. Problem: Determine the domain, range, and sketch the graph of \(f(x) = \sqrt{x^2 - 2x}\). Step 1: The function under the square root must be non-negative: $$x^2 - 2x \geq 0$$ Step 2: Factor the expression: $$x(x-2) \geq 0$$ Step 3: Find the critical points: \(x=0\) and \(x=2\). Step 4: Analyze intervals: - For \(x < 0\): both \(x < 0\) and \(x-2 < 0\), product \(> 0\) since negative * negative = positive. - For \(0 \leq x \leq 2\): \(x \geq 0\), \(x-2 \leq 0\), product \(\leq 0\) actually \(\leq 0\) but product is negative or zero. - For \(x > 2\): both positive, product positive. Step 5: Since \(x(x-2) \geq 0\), domain is \(( -\infty, 0 ] \cup [ 2, \infty )\). Step 6: Determine the range: Because \(f(x)\) is a square root, values are \(\geq 0\). Minimum value is 0 at \(x=0\) and \(x=2\); No maximum since for large \(|x|\), \(f(x) \approx |x|\). Thus range is \([0, \infty)\). Answer: Domain: \(( -\infty, 0 ] \cup [ 2, \infty )\), Range: \([0, \infty)\). 3. Problem: Determine if \(f(x) = x^2 - x\) is even, odd, or neither. Step 1: Recall definitions: - Even: \(f(-x) = f(x)\) - Odd: \(f(-x) = -f(x)\) Step 2: Calculate \(f(-x)\): $$f(-x) = (-x)^2 - (-x) = x^2 + x$$ Step 3: Compare: - \(f(-x) = x^2 + x\) - \(f(x) = x^2 - x\) Not equal, so not even. Step 4: Check if odd: Is \(f(-x) = -f(x)\)? \(x^2 + x \stackrel{?}{=} - (x^2 - x) = -x^2 + x\) No, so not odd. Answer: The function is neither even nor odd. 4. Problem: Check continuity of \[ f(x) = \begin{cases} -3x + 4, & x \leq 2 \\ 2, & x > 2 \end{cases} \] Step 1: To check continuity at \(x=2\), verify: - \(f(2)\): Using first case since \(2 \leq 2\), \(f(2) = -3(2) + 4 = -6 +4 = -2\) - Limit from left: $$\lim_{x \to 2^-} f(x) = -3(2) + 4 = -2$$ - Limit from right: $$\lim_{x \to 2^+} f(x) = 2$$ Step 2: Since left limit \(\neq\) right limit, function is not continuous at \(x=2\). Answer: \(f(x)\) is not continuous at \(x=2\). 5. Problem: Prove using definition of limit that \(\lim_{x \to 3} (x^2 - 2x) = 3\). Step 1: Definition: For every \(\epsilon > 0\), find \(\delta > 0\) such that if \(|x -3| < \delta\) then \(|f(x) - 3| < \epsilon\). Step 2: Compute: \[|f(x) - 3| = |x^2 - 2x - 3| = |(x - 3)(x + 1)|\] Step 3: To bound \(|x + 1|\), restrict \(\delta \leq 1\) so that if \(|x - 3| < 1\), then \(2 < x < 4\), thus \(|x+1| \leq 5\). Step 4: So \[|f(x) - 3| = |x-3||x+1| < 5|x-3|\] Step 5: To have \(|f(x) - 3| < \epsilon\), require \(5|x-3| < \epsilon \implies |x-3| < \epsilon / 5\). Step 6: Choose \(\delta = \min(1, \epsilon/5)\). Answer: By definition of limit, \(\lim_{x \to 3} (x^2 -2x) = 3\). 6. Problem: Find \(\lim_{x \to 0} \frac{\cos 2x -1}{\sin x}\). Step 1: Note that both numerator and denominator go to 0 as \(x \to 0\). Step 2: Use Taylor expansions: \(\cos 2x \approx 1 - 2x^2\) \(\sin x \approx x\) Step 3: Substitute: \[\frac{\cos 2x -1}{\sin x} \approx \frac{(1 - 2x^2) -1}{x} = \frac{-2x^2}{x} = -2x\] Step 4: As \(x \to 0\), \(-2x \to 0\). Answer: The limit is 0.