PCC SLC Math Resources

Problem Set 1

1. \(\begin{aligned}[t] g(\highlight{7})\amp=8-4(\highlight{7})\\ \amp=8-28\\ \amp=-20 \end{aligned}\)
2. \(\begin{aligned}[t] f(\highlight{-3})\amp=-(\highlight{-3})^2+5(\highlight{-3})+12\\ \amp=-9-15+12\\ \amp=-12 \end{aligned}\)
3. \(\begin{aligned}[t] h(\highlight{33})\amp=-\sqrt{\frac{\highlight{33}-25}{2}}\\ \amp=-\sqrt{\frac{8}{2}}\\ \amp=-\sqrt{4}\\ \amp=-2 \end{aligned}\)
4. \(\begin{aligned}[t] t(\highlight{22})\amp=17\\ \end{aligned}\)
5. \(\begin{aligned}[t] y(\highlight{-6})\amp=\abs{-9-(\highlight{-6})}+3\\ \amp=\abs{-3}+3\\ \amp=3+3\\ \amp=6 \end{aligned}\)
6. \(\begin{aligned}[t] z(\highlight{-1})\amp=\frac{\highlight{-1}+6}{(\highlight{-1})^2+1}\\ \amp=\frac{5}{2} \end{aligned}\)
7. \(\begin{aligned}[t] k(\highlight{4})\amp=3\sqrt{21-\highlight{4}^2}\\ \amp=3\sqrt{21-16}\\ \amp=3\sqrt{5} \end{aligned}\)

Problem Set 2

1. Solve \(f(x)=12\) where \(f(x)=4-x\text{.}\)
   \begin{align*} f(x)\amp=12\\ 4-x\amp=12\\ 4-x\subtractright{4}\amp=12\subtractright{4}\\ -x\amp=8\\ \multiplyleft{-1}-x\amp=\multiplyleft{-1}8\\ x\amp=-8 \end{align*}
   The solution set is \(\{-8\}\text{.}\)
2. Solve \(h(t)=100\) where \(h(t)=t^2-21\text{.}\)
   \begin{align*} h(t)\amp=100\\ t^2-21\amp=100\\ t^2-21\addright{21}\amp=100\addright{21}\\ t^2\amp=121\\ t\amp=\pm\sqrt{121}\\ t\amp=\pm 11 \end{align*}
   The solution set is \(\{-11,11\}\text{.}\)
3. Solve \(p(x)=-14\) where \(p(x)=6-\abs{x}\text{.}\)
   \begin{align*} p(x)\amp=-14\\ 6-\abs{x}\amp=-14\\ 6-\abs{x}\subtractright{6}\amp=-14\subtractright{6}\\ -\abs{x}\amp=-20\\ \multiplyleft{-1}-\abs{x}\amp=\multiplyleft{-1}-20\\ \abs{x}\amp=20\\ x\amp=\pm 20 \end{align*}
   The solution set is \(\{-20,20\}\text{.}\)
4. Solve \(w(y)=10\) where \(w(y)=y^2-y-2\text{.}\)
   \begin{align*} w(y)\amp=10\\ y^2-y-2\amp=10\\ y^2-y-2\subtractright{10}\amp=10\subtractright{10}\\ y^2-y-12\amp=0\\ (y-4)(y+3)\amp=0 \end{align*}
   \begin{align*} y-4\amp=0\amp\amp\text{ or }\amp y+3\amp=0\\ y-4\addright{4}\amp=0\addright{4}\amp\amp\text{ or }\amp y+3\subtractright{3}\amp=0\subtractright{3}\\ y\amp=4\amp\amp\text{ or }\amp y\amp=-3 \end{align*}
   The solution set is \(\{-3,4\}\text{.}\)
5. Solve \(r(t)=s(t)\) where \(r(t)=\frac{5}{7}t-3\) and \(s(t)=\frac{2}{3}t+\frac{11}{21}\text{.}\)
   \begin{align*} r(t)\amp=s(t)\\ \frac{5}{7}t-3\amp=\frac{2}{3}t+\frac{11}{21}\\ \multiplyleft{21}(\frac{5}{7}t-3)\amp=\multiplyleft{21}(\frac{2}{3}t+\frac{11}{21})\\ 15t-63\amp=14t+11\\ 15t-63\addright{63}\amp=14t+11\addright{63}\\ 15t\amp=14t+74\\ 15t\subtractright{14t}\amp=14t+74\subtractright{14t}\\ t\amp=74 \end{align*}
   The solution set is \(\{74\}\text{.}\)
6. Solve \(g(x)=y(x)\) where \(q(x)=6-4x^2\) and \(y(x)=(3-x)(8+4x)\text{.}\)
   \begin{align*} g(x)\amp=y(x)\\ 6-4x^2\amp=(3-x)(8+4x)\\ 6-4x^2\amp=24+4x-4x^2\\ 6-4x^2\addright{4x^2}\amp=24+4x-4x^2\addright{4x^2}\\ 6\amp=24+4x\\ 6\subtractright{24}\amp=24+4x\subtractright{24}\\ -18\amp=4x\\ \divideunder{-18}{4}\amp=\divideunder{4x}{4}\\ -\frac{9}{2}\amp=x \end{align*}
   The solution set is \(\{-\frac{9}{2}\}\text{.}\)

Problem Set 3

1. \(f(4)=2\)
2. \(f(2)=2\)
3. \(f(5)\) is not defined.
4. \(f(1)=5\)

1. The solution set is \(\{-1,3\}\text{.}\)
2. The solution set is \(\{-4,-2,0,2,4\}\text{.}\)
3. The solution set is \(\emptyset\text{.}\)
4. The solution set is \(\{-3,1\}\)

1. Simplify \(f(x-7)\) for the function \(f(x)=3x+12\text{.}\)
   \begin{align*} f(\highlight{x-7})\amp=3(\highlight{x-7})+12\\ \amp=3x-21+12\\ \amp=3x-9 \end{align*}
2. Simplify \(g(x)+9\) for the function \(g(x)=14-7x\text{.}\)
   \begin{align*} g(x)\highlightr{+9}\amp=14-7x\highlightr{+9}\\ \amp=23-7x \end{align*}
3. Simplify \(h(5-2t)\) for the function \(h(t)=\sqrt{3-t^2}\text{.}\)
   \begin{align*} h(\highlight{5-2t})\amp=\sqrt{3-(\highlight{5-2t})^2}\\ \amp=\sqrt{3-(5-2t)(5-2t)}\\ \amp=\sqrt{3-(25-20t+4t^2)}\\ \amp=\sqrt{3-25+20t-4t^2}\\ \amp=\sqrt{-4t^2+20t-22} \end{align*}
4. Simplify \(k(x+4)+3\) for the function \(k(x)=2x^2-3\text{.}\)
   \begin{align*} k(\highlight{x+4})\highlightr{+3}\amp=2(\highlight{x+4})^2-3\highlightr{+3}\\ \amp=2(x+4)(x+4)\\ \amp=2(x^2+8x+16)\\ \amp=2x^2+16x+32 \end{align*}
5. Simplify \(r(\sqrt{t-4})\) for the function \(r(t)=3-7t^2\text{.}\)
   \begin{align*} r(\highlight{\sqrt{t-4}})\amp=3-7(\highlight{\sqrt{t-4}})^2\\ \amp=3-7(t-4)\\ \amp=3-7t+28\\ \amp=31-7t \end{align*}
6. Simplify \(y(t+5)-y(t-3)\) for the function \(y(t)=3t-16\text{.}\)
   \begin{align*} y(\highlight{t+5})-y(\highlightb{t-3})\amp=(3(\highlight{t+5})-16)-(3(\highlightb{t-3})-16)\\ \amp=3t+15-16-(3t-9-16)\\ \amp=3t-1-(3t-25)\\ \amp=3t-1-3t+25\\ \amp=24 \end{align*}
7. Simplify \(w(x-2)+w(2-x)+2\) for the function \(w(x)=5-x\text{.}\)
   \begin{align*} w(\highlight{x-2})+w(\highlightb{2-x})\highlightr{+2}\amp=(5-(\highlight{x-2}))+(5-(\highlightb{2-x}))\highlightr{+2}\\ \amp=5-x+2+5-2+x+2\\ \amp=12 \end{align*}
8. Simplify \(s(t-8)\) for the function \(s(t)=\frac{t+8}{t}\text{.}\)
   \begin{align*} s(\highlight{t-8})\amp=\frac{\highlight{t-8}+8}{\highlight{t-8}}\\ \amp=\frac{t}{t-8} \end{align*}
9. Simplify \(3u(2x)\) for the function \(u(x)=-x^2\text{.}\)
   \begin{align*} \highlightg{3}u(\highlight{2x})\amp=\highlightg{3\cdot}-(\highlight{2x})^2\\ \amp=-3 \cdot 4x^2\\ \amp=-12x^2 \end{align*}
10. Simplify \(\frac{1}{2}h(2t-7)-8\) for the function \(h(t)=8t-12\text{.}\)
    \begin{align*} \highlightg{\frac{1}{2}}h(\highlight{2t-7})\highlightr{-8}\amp=\highlightg{\frac{1}{2}}(8(\highlight{2t-7})-12)\highlightr{-8}\\ \amp=\frac{1}{2}(16t-56)-20\\ \amp=8t-28-20\\ \amp=8t-48 \end{align*}

1. To determine the domain of \(y(x)=\sqrt{15-x}\text{,}\) we begin by noting that we cannot take the square root of a negative number (at least over the real numbers). This gives us the following.

   \begin{align*} 15-x \amp\ge 0\\ 15-x\subtractright{15} \amp\ge 0\subtractright{15}\\ -x \amp\ge -15\\ \multiplyleft{-1}-x \amp\le \multiplyleft{-1}-15\\ x \amp\le 15 \end{align*}

   The domain of \(y\) is \((-\infty,15)\)

2. To determine the domain of \(f(t)=\sqrt[3]{t^2-9}\text{,}\) we begin by noting that the polynomial expression \(t^2-9\) is defined for all real numbers as is the cube root function. So the domain of \(f\) is \((-\infty,\infty)\text{.}\)

3. To determine the domain of \(w(x)=\frac{x-7}{x-12}\text{,}\) we begin by noting that we cannot divide by zero. This gives us the following.

   \begin{align*} x-12 \amp\ne 0\\ x-12\addright{12} \amp\ne 0\addright{12}\\ x \amp\ne 12 \end{align*}

   The domain of \(w\) is \((-\infty,12) \cup (12,\infty)\text{.}\)

4. To determine the domain of \(g(x)=\frac{x+3}{x^2+8x+15}\text{,}\) we begin by noting that we cannot divide by zero. This gives us the following.

   \begin{align*} x^2+8x+15 \amp\ne 0\\ (x+3)(x+5) \amp\ne 0 \end{align*}
   \begin{align*} x+3 \amp\ne 0 \amp\amp\text{ and }\amp x+5 \amp\ne 0\\ x+3\subtractright{3} \amp\ne 0\subtractright{3} \amp\amp\text{ and }\amp x+5\subtractright{5} \amp\ne 0\subtractright{5}\\ x \amp\ne -3 \amp\amp\text{ and }\amp x \amp\ne -5 \end{align*}

   The domain of \(g\) is \((-\infty,-5) \cup (-5,-3) \cup (-3,\infty)\text{.}\)

5. To determine the domain of \(r(t)=t^2-3t+9\text{,}\) we begin by noting that \(r\) is a polynomial function and polynomial functions are defined for all real numbers. So the domain of \(r\) is \((-\infty,\infty)\text{.}\)

6. To determine the domain of \(k(t)=\frac{t^2+16}{t^2+16}\text{,}\) we begin by noting that we cannot divide by zero. But \(t^2+16\) is positive for all real number values of \(t\text{,}\) so no value of \(t\) will cause division by zero. So the domain of \(k\) is \((-\infty,\infty)\text{.}\)

Problem Set 1

1. The solution set is \(\{x \mid -1 \leq x \lt 0 \text{ or } 2 \lt x \leq 5\}\text{.}\)

   The solution set is \([-1,0) \cup (2,5]\text{.}\)

2. The solution set is \(\{x \mid x \leq 0 \text{ or } x \geq 4\}\text{.}\)

   The solution set is \((-\infty,0] \cup [4,\infty)\text{.}\)

3. The solution set is \(\{\}\text{.}\)

   The solution set is \(\emptyset\text{.}\)

4. The solution set is \(\{x \mid x \leq 0 \text{ or } x \gt 2\}\text{.}\)

   The solution set is \((-\infty,0] \cup (2,\infty)\text{.}\)

Problem Set 2

1. The solution set is \(\{x \mid -5 \leq x \leq 1\}\text{.}\)

   The solution set is \([-5,1]\text{.}\)

2. The solution set is \(\{x \mid x \lt -5 \text{ or } x \gt 1\}\text{.}\)

   The solution set is \((-\infty,-5) \cup (1,\infty)\text{.}\)

1. We are tasked to solve \(2t-7 \gt 11 \text{ or } 3-t \lt 6\text{.}\) Let's do so.

   \begin{align*} 2t-7 \amp\gt 11 \amp\amp\text{ or }\amp 3-t \amp\lt 6\\ 2t-7\addright{7} \amp\gt 11\addright{7} \amp\amp\text{ or }\amp 3-t\subtractright{3} \amp\lt 6\subtractright{3}\\ 2t \amp\gt 18 \amp\amp\text{ or }\amp -t \amp\lt 3\\ \divideunder{2t}{2} \amp\gt \divideunder{18}{2} \amp\amp\text{ or }\amp \multiplyleft{-1}-t \amp\gt \multiplyleft{-1}3\\ t \amp\gt 9 \amp\amp\text{ or }\amp t \amp\gt -3 \end{align*}

   Since the first set is completely contained within the second, the solution set is the second set, \((-3,\infty)\text{.}\)

2. We are tasked to solve \(-\frac{4}{5}x+\frac{3}{2} \leq 1 \text{ or } 2x-7 \lt -15\text{.}\) We begin by clearing the fraction from the first inequality.

   \begin{align*} -\frac{4}{5}x+\frac{3}{2} \amp\leq 1\\ \multiplyleft{10}(-\frac{4}{5}x+\frac{3}{2}x) \amp\leq \multiplyleft{10}1\\ -8x+15 \amp\leq 10 \end{align*}

   We now solve both inequalities.

   \begin{align*} -8x+15 \amp\leq 10 \amp\amp\text{ or }\amp 2x-7 \amp\lt -15\\ -8x+15\subtractright{15} \amp\leq 10\subtractright{15} \amp\amp\text{ or }\amp 2x-7\addright{7} \amp\lt -15\addright{7}\\ -8x \amp\leq -5 \amp\amp\text{ or }\amp 2x \amp\lt -8\\ \divideunder{-8x}{-8} \amp\geq \divideunder{-5}{-8} \amp\amp\text{ or }\amp \divideunder{2x}{2} \amp\lt \divideunder{-8}{2}\\ x \amp\geq \frac{5}{8} \amp\amp\text{ or }\amp x \amp\lt -4 \end{align*}

   The solution set is \((-\infty,-4) \cup [\frac{5}{8}\infty)\text{.}\)

3. We are tasked to solve \(7 \leq 4y+1 \lt 13\text{.}\) Let's do so.

   \begin{alignat*}{2} 7 \amp\leq 4y+1 \amp\amp\lt 13\\ 7\subtractright{1} \amp\leq 4y+1\subtractright{1} \amp\amp\lt 13\subtractright{1}\\ 6 \amp\leq 4y \amp\amp\lt 12\\ \divideunder{6}{4} \amp\leq \divideunder{4y}{4} \amp\amp\lt \divideunder{12}{4}\\ \frac{3}{2} \amp\leq y \amp\amp\lt 3 \end{alignat*}

   The solution set is \([\frac{3}{2},3)\text{.}\)

4. We are tasked to solve \(14 \gt 2-3x \gt 11\text{.}\) Let's do so.

   \begin{alignat*}{2} 14 \amp\gt 2-3x \amp\amp\gt 11\\ 14\subtractright{2} \amp\gt 2-3x\subtractright{2} \amp\amp\gt 11\subtractright{2}\\ 12 \amp\gt -3x \amp\amp\gt 9\\ \divideunder{12}{-3} \amp\lt \divideunder{-3x}{-3} \amp\amp\lt \divideunder{9}{-3}\\ -4 \amp\lt x \amp\amp\lt -3 \end{alignat*}

   The solution set is \((-4,-3)\text{.}\)

1. Our opportunity, not to be passed, is to determine the solution set for \(2\abs{3x}-12 \le -6\text{.}\) We begin by isolating the absolute value expression.

   \begin{align*} 2\abs{\frac{x}{3}}-12 \amp\le -6\\ 2\abs{\frac{x}{3}}-12\addright{12} \amp\le -6\addright{12}\\ 2\abs{\frac{x}{3}} \amp\le 6\\ \divideunder{2\abs{\frac{x}{3}}}{2} \amp\le \divideunder{6}{2}\\ \abs{\frac{x}{3}} \amp\le 3 \end{align*}

   We now write and solve an equivalent compound equality that does not include an absolute value expression.

   \begin{alignat*}{2} -3 \amp\le \frac{x}{3} \amp \amp\le 3\\ \multiplyleft{3}-3 \amp\le \multiplyleft{3}\frac{x}{3} \amp \amp\le \multiplyleft{3}3\\ -9 \amp\le x \amp \amp\le 9 \end{alignat*}

   The solution set to the given inequality is \([-9,9]\text{.}\)

2. Our opportunity, which we will surely seize, is to determine the solution set for \(5-6\abs{2x+1}=17\text{.}\) We begin by isolating the absolute value expression.

   \begin{align*} 5-6\abs{2x+1}\amp=17\\ 5-6\abs{2x+1}\subtractright{5}\amp=17\subtractright{5}\\ -6\abs{2x+1}\amp=12\\ \divideunder{-6\abs{2x+1}}{-6}\amp=\divideunder{12}{-6}\\ \abs{2x+1}\amp=-2 \end{align*}

   We observe that there are no numbers whose absolute value is \(-2\) and conclude that the given equation has no solutions.

3. We have the pleasure of determining the solution set for \(\abs{y-9}=\abs{12-2y}\text{.}\) We begin by observing that the only way the two absolute value expressions can be equal is if the expressions inside the absolute value bars are either equal or opposites. This give us the following.

   \begin{align*} y-9\amp=12-2y \amp\amp\text{ or }\amp y-9\amp=-(12-2y)\\ y-9\amp=12-2y \amp\amp\text{ or }\amp y-9\amp=-12+2y\\ y-9\addright{9}\amp=12-2y\addright{9} \amp\amp\text{ or }\amp y-9\addright{9}\amp=-12+2y\addright{9}\\ y\amp=21-2y \amp\amp\text{ or }\amp y\amp=-3+2y\\ y\addright{2y}\amp=21-2y\addright{2y} \amp\amp\text{ or }\amp y\subtractright{2y}\amp=-3+2y\subtractright{2y}\\ 3y\amp=21 \amp\amp\text{ or }\amp -y\amp=-3\\ \divideunder{3y}{3}\amp=\divideunder{21}{3} \amp\amp\text{ or }\amp \multiplyleft{-1}-y\amp=\multiplyleft{-1}-3\\ y\amp=7 \amp\amp\text{ or }\amp y\amp=3 \end{align*}

   The solution set is \(\{3,7\}\text{.}\)

4. Let's determine the solution set for \(-\abs{\frac{8-2x}{3}}+15 \le 12\text{.}\) We begin by isolating the absolute value expression.

   \begin{align*} -\abs{\frac{8-2x}{3}}+15 \amp\le 12\\ -\abs{\frac{8-2x}{3}}+15\subtractright{15} \amp\le 12\subtractright{15}\\ -\abs{\frac{8-2x}{3}} \amp\le -3\\ \multiplyleft{-1}-\abs{\frac{8-2x}{3}} \amp\ge \multiplyleft{-1}-3\\ \abs{\frac{8-2x}{3}} \amp\ge 3 \end{align*}

   We now write and solve an equivalent compound inequality that does not include an absolute value expression

   \begin{align*} \frac{8-2x}{3} \amp\le -3 \amp\amp\text{or}\amp \frac{8-2x}{3} \amp\ge 3\\ \multiplyleft{3}\frac{8-2x}{3} \amp\le \multiplyleft{3}-3 \amp\amp\text{or}\amp \multiplyleft{3}\frac{8-2x}{3} \amp\ge \multiplyleft{3}3\\ 8-2x \amp\le -9 \amp\amp\text{or}\amp 8-2x \amp\ge 9\\ 8-2x\subtractright{8} \amp\le -9\subtractright{8} \amp\amp\text{or}\amp 8-2x\subtractright{8} \amp\ge 9\subtractright{8}\\ -2x \amp\le -17 \amp\amp\text{or}\amp -2x \amp\ge 1\\ \divideunder{-2x}{-2} \amp\ge \divideunder{-17}{-2} \amp\amp\text{or}\amp \divideunder{-2x}{-2} \amp\le \divideunder{1}{-2}\\ x \amp\ge \frac{17}{2} \amp\amp\text{or}\amp x \amp\le -\frac{1}{2} \end{align*}

   The solution set to the given inequality is \((-\infty,-\frac{1}{2}] \cup [\frac{17}{2},\infty)\text{.}\)

Problem 1

The population of Portland grew by 56,607 people over a course of 6 years which gives us the following slope.

The population on January 1, 2010 is the \(y\)-coordinate of the \(y\)-intercept, so using the linear form \(y=mt+b\) we have the following.

We can now evaluate \(P(9.5)\text{.}\)

Assuming this last figure to be contextually exact, this tells us that the population of Portland halfway through 2019 wil be 672,582.

Problem 2

1. The vertex tells us that \(1.9\) seconds after the ball was thrown, the vertical displacement between the ball and the floor was \(64\) feet and that this was the greatest vertical displacement between the ball and the floor.

2. The solution set to \(v(t) \geq 30\) is approximately \([.4,3.3]\text{.}\) Assuming the endpoints to be exact, this tells us that the vertical displacement between the ball and the floor was at least 30 feet from \(.4\) seconds after the ball was thrown until 3.3. seconds after the ball was thrown.

3. The endpoint \((3.78,6.2)\) tells us that (presumably) someone caught the ball \(6.2\) feet above the floor \(3.78\) seconds after the ball was thrown.

Problem 3

We begin by making the following observation.

So, if we wanted to convert, say, \(288 \,\text{in}^2\) to square feet, we could make the following calculation.

From this, we can infer that

Now we have the following.

The function value just determined tells us that \(48\) square inches is equivalent to one-third square feet.