Precalculus: Lecture Notes

1. By definition of \(f\text{:}\)

   \begin{equation*} \begin{array}{c||c|c|c|c|c|c|c} n\amp-2\amp-1\amp0\amp1\amp4\amp100\amp111\\ \hline f(n)\amp \sin(-2\pi)=0\amp \cos (-\pi)=-1 \amp \sin(0)=0\amp \cos (\pi)=-1\amp 0 \amp0 \amp-1\\ \end{array} \end{equation*}

2. Observe:

   \begin{equation*} f(n)=\left\{ \begin{array}{ll} 0\amp\text{if } n \text{ is even}\\ -1\amp\text{if }n \text{ is odd}\\ \end{array} \right. \end{equation*}

3. By definition, the domain of the function \(f\) is \(\mathbb{Z}\text{,}\) the set of all integers.

   We observe that the range the function \(f\) is the set \(\{-1,0\}\text{.}\)

4. See Figure 5.35.

   

1. We read from the graph:

   \begin{equation*} \begin{array}{c||c|c|c|c|c} x\amp-2\amp-1\amp0\amp1\amp2\\ \hline f(x)\amp\text{Not defined}\amp 2\amp 0\amp 0.5\amp -1\\ \end{array} \end{equation*}

   .

2. See Figure 5.37.

   

3. See Figure 5.38.

   

1. Let \(x\) represent the ticket price in dollars and let \(A(x)\) represent the average attendance at the price \(x\text{.}\) Observe that, since for each dollar the ticket price is lowered the average attendance increases by 50, the attendance with the ticket price of \(x\) dollars is given by

   \begin{equation*} A(x)=5500+50(75-x)=5500+3750-50x=9250-50x. \end{equation*}

   The revenue is given as the product of the ticket price and the number of the people in the attendance.

   It follows that the function

   \begin{equation*} R(x)=x\cdot A(x)=x(9250-50x)=-50x^2+9250x \end{equation*}

   models the revenue \(R(x)\) in terms of ticket price \(x\text{.}\)

2. Note that \(R(x)\) is a quadratic function and that its graph is part of a parabola that opens downward. Thus the maximum revenue is achieved at the vertex of the parabola. The first coordinate of the vertex is given by

   \begin{equation*} x=-\frac{9250}{2\cdot (-50)}=92.50 \end{equation*}

   Therefore the ticket price of \(x=\$92.50\) will yield the maximum revenue.

   Note: From \(R(x)=-50x^2+9250x=50x(185-x)\) we obtain that the maximum revenue is \(R(92.50)=50\cdot 92.50\cdot(185-92.50)=4625\cdot 92.50=427812.50\) dollars.

   Also, at the price of \(\$92.50\) there will be \(A(92.50)=9250-50\cdot 92.50=4625\) spectators in the arena.

1. The domain of the function \(f\) is given by

   \begin{equation*} \{x\in \mathbb{R}: 1-x\geq 0\}=\{x\in \mathbb{R}: x\leq 1\}=(\infty, -1]. \end{equation*}

2. If we take \(s(x)=1-x\text{,}\) \(r(x)=\sqrt{x}\text{,}\) \(q(x)=e^x\text{,}\) and \(p(x)=x+3\) then

   \begin{equation*} (p\circ q \circ r\circ s)(x)=p(q(r(s(x))))=p(q(r(1-x)))=p(q(\sqrt{1-x}))=p(e^{\sqrt{x}})=e^{\sqrt{1-x}}+3=f(x). \end{equation*}

3. To find the inverse function of the function \(f\) we apply the following procedure.

   Let \(y=e^{\sqrt{1-x}}+3\text{.}\) We interchange the roles of “\(x\)” and “\(y\)” to obtain

   \begin{equation*} x=e^{\sqrt{1-y}}+3 \end{equation*}

   and then we solve this expression for “\(y\)”.

   Thus

   \begin{equation*} x=e^{\sqrt{1-y}}+3\Leftrightarrow x-3=e^{\sqrt{1-y}}\Leftrightarrow \ln(x-3)=\sqrt{1-y} \end{equation*}
   \begin{equation*} \Leftrightarrow ((\ln (x-3))^2=1-y\Leftrightarrow y=1-((\ln (x-3))^2. \end{equation*}

   The inverse function of the function \(f\) is given by \(f^{-1}(x)=1-((\ln (x-3))^2\text{.}\)

1. Let \(t=5^x\text{.}\) With this substitution the given equation becomes \(t^2-t-12=0\text{.}\)

   We solve this quadratic equation:

   \begin{equation*} t^2-t-12=0\Leftrightarrow (t-4)(t+3)=0\Leftrightarrow (t=4 \text{ or }t=-3). \end{equation*}

   Since \(t=5^x>0\text{,}\) we reject the negative value \(t=-3\text{.}\)

   It follows

   \begin{equation*} 5^x=4\Leftrightarrow x=\log_54. \end{equation*}

2. We use the fact that \(\log(10x)=\log 10+\log x=1+\log x\) and \(\log(100x)=\log 100+\log x=2+\log x\) to obtain

   \begin{equation*} \left[\log (10x)\right]\cdot \left[\log(100x)\right]=(\log x)^2\Leftrightarrow (1+\log x)\cdot(2+\log x)=(\log x)^2\Leftrightarrow \end{equation*}
   \begin{equation*} 2+3\log x+(\log x)^2=(\log x)^2\Leftrightarrow 2+3\log x=0\Leftrightarrow \log x=-\frac{2}{3}\Leftrightarrow x=10^{-\frac{2}{3}}. \end{equation*}

3. We use the fact that \(\sin 2x=2\sin x\cos x\) to obtain

   \begin{equation*} \sin 2x-\sqrt{2}\cdot \sin x=0\Leftrightarrow 2\sin x\cos x-\sqrt{2}\cdot \sin x=0\Leftrightarrow \sin x\cdot (2\cos x-\sqrt{2})=0. \end{equation*}

   Hence

   \begin{equation*} \sin 2x-\sqrt{2}\cdot \sin x=0\Leftrightarrow (\sin x=0 \text{ or } 2\cos x-\sqrt{2}=0). \end{equation*}

   Finally, we observe that since, \(x\in[0,2\pi )\text{,}\)

   \begin{equation*} \sin x=0 \Leftrightarrow x=0 \text{ or } x=\pi \end{equation*}

   and

   \begin{equation*} 2\cos x-\sqrt{2}=0\Leftrightarrow \cos x=\frac{\sqrt{2}}{2}\Leftrightarrow x=\frac{\pi}{4} \text{ or }x=\frac{7\pi}{4}. \end{equation*}

   Therefore, for \(x\in[0,2\pi )\text{,}\)

   \begin{equation*} \sin 2x-\sqrt{2}\cdot \sin x=0\Leftrightarrow x\in \left\{0,\frac{\pi}{4},\pi, \frac{7\pi}{4}\right\}. \end{equation*}

1. We note that the centre of the circle is at the point \((-2,3)\) and that its radius is \(r=2\text{.}\) Hence the equation of the circle is

   \begin{equation*} (x+2)^2+(y-3)^2=4. \end{equation*}

2. The first coordinate of the point \(P\) is \(x=-3\text{.}\) We find the second coordinate, \(y\text{,}\) by using the fact that the point is on the circle:

   \begin{equation*} (-3+2)^2+(y-3)^2=4\Leftrightarrow 1+(y-3)^2=4\Leftrightarrow (y-3)^2=3\Leftrightarrow y-3=\pm\sqrt{3}. \end{equation*}

   Since the point \(P\) is below the centre of the circle, it follows that \(y\lt 3\text{,}\) i.e \(y=3-\sqrt{3}\text{.}\) Hence, \(P=(-3,3-\sqrt{3})\text{.}\)

3. The line \(L\) is perpendicular to the line through \(P\) and the centre of the circle. Hence its slope is \(m=-\frac{1}{\sqrt{3}}\text{.}\)

   The line is given by

   \begin{equation*} y-(3-\sqrt{3})=-\frac{1}{\sqrt{3}}\cdot (x+3). \end{equation*}

1. The total length of the fence is given by \(L=2x+3y\text{.}\) From the fact that the area of the plot is \(2000 \textrm{m}^2\) we conclude that

   \begin{equation*} 2000=xy\Rightarrow x=\frac{2000}{y}. \end{equation*}

   It follows that the total length \(L\) of fence used as a function of \(y\) is given by

   \begin{equation*} L=L(y)=2\cdot \frac{2000}{y}+3y=\frac{4000}{y}+3y=\frac{4000+3y^2}{y}. \end{equation*}

   This is a rational function.

2. Since \(y\) represents the length of one side of the plot, we have that the domain of the function \(L=L(y)\) is the set of all positive real numbers, \((0,\infty )\text{.}\)

3. From \(L(y) = \frac{4000}{y}+3y\) we observe that if \(y\) is large then \(\frac{4000}{y}\) is small and \(L(y)\approx 3y\) is large. If \(y\) is small then \(L(y)\approx \frac{4000}{y}\) is large.

4. Observe that \(L(y)>0\) for \(y>0\) and that for \(y\) very large or very small, \(L(y)\) is very large. Hence the graph of the function \(L\) may look like one on Figure 5.41. The graph clearly suggests that there must be a value of\(y\) that will yield the minimum length of the fence \(L\text{.}\)

   

1. We note that, for \(|h|\not= 0\text{,}\)

   \begin{equation*} \frac{B_1(1+h)-B_1(1)}{h}=\frac{(1+h)-\frac{1}{2}-(1-\frac{1}{2})}{h}=\frac{1+h-\frac{1}{2}-1+\frac{1}{2}}{h}=\frac{h}{h}=1=B_0(1). \end{equation*}

2. From, for \(|h|\not= 0\text{,}\)

   \begin{equation*} \begin{array}{rcl} \frac{B_2(1+h)-B_2(1)}{h}\amp =\amp \frac{(1+h)^2-(1+h)+\frac{1}{6}-(1^2-1+\frac{1}{6})}{h}\\ \amp =\amp \frac{1+2h+h^2-1-h+\frac{1}{6}-\frac{1}{6}}{h}\\ \amp =\amp \frac{1+2h+h^2-1-h}{h}=\frac{h(1+h)}{h}=1+h\\ \end{array} \end{equation*}

   and \(B_{1}(1)=1-\frac{1}{2}=\frac{1}{2}\) we conclude that, for a small value of \(|h|\not= 0\text{,}\)

   \begin{equation*} \frac{B_2(1+h)-B_2(1)}{h}=1+h\approx 1=2\left(1-\frac{1}{2}\right)=2B_1(1). \end{equation*}

1. Observe, \(\ln 8=\ln (2^3)=3\ln 2\approx 3\cdot 0.6=1.8\text{.}\)

2. Observe, \(\ln 25=\ln (5^2)=2\ln 5\approx 2\cdot 1.6=3.2\text{.}\)

3. Observe, \(\ln 10=\ln (2\cdot 5)=\ln 2+\ln 5\approx 0.6+1.6=2.2\text{.}\)

4. Observe, \(ln 2.5=\ln \frac{25}{10}=\ln 25-\ln 10\approx 3.2-2.2=1\text{.}\)

5. Observe, \(\ln 40=\ln (8\cdot 5)=\ln 8+\ln 5\approx 1.8+1.6=3.4\text{.}\)

6. Observe, \(\ln 50=\ln (25\cdot 2)=\ln 25+\ln 2\approx 3.2+ 0.6=3.8\text{.}\)

7. Observe,

   \begin{equation*} \log_25=\frac{\ln 5}{\ln 2}\approx \frac{1.6}{0.6}\approx 2.66\text{.} \end{equation*}

Note that \(\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}\text{.}\) It follows that the third angle in the triangle with the vertices “Laura,” “Sign,” and “Ali” equals \(\frac{\pi}{2}\) and that this triangle is a right triangle with the right angle at the vertex “Sign” and the hypothenuse of the length \(10\text{.}\)

Let \(x\) be the distance between Laura and the sign, Then

See Figure 5.42

1. Since \(\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}\text{,}\) it follows

   \begin{equation*} \cos \frac{\pi}{12}=\cos \left( \frac{\pi}{4}-\frac{\pi}{6}\right)=\cos \frac{\pi}{4}\cos \frac{\pi}{6}+\sin \frac{\pi}{4}\sin \frac{\pi}{6}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{equation*}

2. Since \(\frac{7\pi}{12}=\frac{\pi}{2}+\frac{\pi}{12}\text{,}\) it follows

   \begin{equation*} \sin \frac{7\pi}{12}=\sin \left( \frac{\pi}{2}+\frac{\pi}{12}\right)=\sin \frac{\pi}{2}\cos \frac{\pi}{12}+\cos \frac{\pi}{2}\sin \frac{\pi}{12}=1\cdot\frac{\sqrt{6}+\sqrt{2}}{4}+0\cdot\sin \frac{\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{equation*}

3. By definition, \(\tan \frac{\pi}{12}=\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}}\text{.}\)

   From \(\cos \frac{\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}\) it follows

   \begin{equation*} \sin^2\left( \frac{\pi}{12}\right)=1-\cos^2\left(\frac{\pi}{12}\right)=1-\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^2=1-\frac{6+2\sqrt{12}+2}{16} \end{equation*}
   \begin{equation*} =1-\frac{8+4\sqrt{3}}{16}=1-\frac{2+\sqrt{3}}{4}=\frac{4-2-\sqrt{3}}{4}=\frac{2-\sqrt{3}}{4}. \end{equation*}
   \begin{equation*} =\frac{3-2\sqrt{3}+1}{8}=\frac{(\sqrt{3}-1)^2}{8}. \end{equation*}

   Next we note that \(\frac{\pi}{12}\in\left(0,\frac{\pi}{2}\right)\) which implies that

   \begin{equation*} \sin \frac{\pi }{12}=\sqrt{\frac{(\sqrt{3}-1)^2}{8}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\gt 0. \end{equation*}

   Finally,

   \begin{equation*} \tan \frac{\pi}{12}=\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \end{equation*}

4. By definition, \(\cot \frac{\pi}{12}=\frac{1}{\tan \frac{\pi }{12}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\text{.}\)

1. We observe that the graph of the function \(f\) passes the horizontal line test, i.e. each horizontal line intersects the graph of \(f\) at most one point. Hence, the function \(f\) is one–to–one, which implies that \(f\) has an inverse.

2. Since \(f:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\to [-1,1]\) we have that \(f^{-1}:[-1,1]\to \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}\) Therefore, the domain of the function \(f^{-1}\) is the segment \([-1,1]>\) and its range is the segment \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}\)

3. See Figure 5.44