Week 2 Worksheet

  1. Solve for \(x\) in \(5^{x/2} = 100\).

    Solution

    Take \(\log_5\) of both sides:

    \begin{align*} \log_5\left(5^{x/2}\right) &= \log_5 100 \\ \frac{x}{2} &= \log_5 100 \\ x &= 2\log_5 100 \approx 5.723 \end{align*}

    Note that other forms like \(\log_5 10000\) and \(4 + 2\log_5 4\) are also valid answers.

  2. What is the largest possible domain for the function \(f(x) = \frac{1}{\sqrt{1 - x^2}}\)? What about for \(g(x) = \sqrt{1 - x^2}\)?

    Solution

    For \(f\) to make sense, the denominator must make sense and also be nonzero. For \(\sqrt{1 - x^2}\) to make sense, we need \(1 - x^2 \geq 0\), or equivalently, \(1 \geq x^2\). Looking at the graph of \(x^2\), this happens exactly when \(-1 \leq x \leq 1\). For \(\sqrt{1 - x^2}\) to be nonzero, we need \(1 - x^2 \neq 0\), which means \(x \neq 1\) and \(x \neq -1\). Thus the domain of \(f\) is \(-1 < x < 1\), or in interval notation, \((-1, 1)\).

    For \(g\), we can repeat the same argument, except we don’t need \(\sqrt{1 - x^2}\) to be nonzero, so the domain is \(-1 \leq x \leq 1\), or \([-1, 1]\).

  3. Find the domain for the function \[ f(x) = \sqrt{\frac{x^3 + 2x^2 + x}{x-5}}. \]

    Solution

    For \(f\) to be defined, we need the denominator to be non-zero, and we need the argument to the square root to be non-negative. Both of these can be found by factoring:

    \[ \frac{x^3 + 2x^2 + x}{x-5} = \frac{x(x+1)^2}{x-5} \]

    Points of interest are \(x = -1, 0, 5\). At \(x = -1,0\), we have \(f(x) = \sqrt{0} = 0\), and everything is defined. At \(x = 5\) we are dividing by zero, so \(f(x)\) is undefined. For all other points, we make a table:

    Table 1: Sign analysis for \(\frac{x(x+1)^2}{x+5}\).
    Expression \((-\infty, -1)\) \((-1, 0)\) \((0, 5)\) \((5,\infty)\)
    \(x\) \(-\) \(-\) \(+\) \(+\)
    \((x+1)^2\) \(+\) \(+\) \(+\) \(+\)
    \(x-5\) \(-\) \(-\) \(-\) \(+\)
    \(\frac{x(x+1)^2}{x+5}\) \(+\) \(+\) \(-\) \(+\)

    Putting this all together, our domain is \((-\infty, 0] \cup (5, \infty)\).

  4. Label each of the following functions as even, odd, both, or neither:

    1. \(f(x) = x^5 + x^3 + x\)

      Solution

      Odd. This is because \(x^5\), \(x^3\) and \(x\) are all odd functions, so their sum is also odd.

    2. \(g(x) = 5\)

      Solution

      Even. This is because \(g(x) = g(-x)\) for all \(x\). (Both sides are always 5).

    3. \(h(x) = x^3 + x^2\)

      Solution

      Neither. It is sufficient to find a single point where \(h(x) \neq \pm h(-x)\). In particular, \(h(1) = 2\), but \(h(-1) = 0\).

  5. Give an example of a function which is both even and odd. Do any other examples exist? Justify your answer.

    Solution

    Consider \(f(x) = 0\). This is even, since \(f(-x) = 0 = -f(x)\) for all \(x\). It is also odd, since \(f(-x) = 0 = -0 = -f(x)\) for all \(x\).

    This is the only example. To see why, suppose \(g\) were any other function which were both even and odd. Then for all \(x\), we have both \(g(-x) = g(x)\) and \(g(-x) = -g(x)\). Putting these together, we get

    \begin{align*} g(x) &= -g(x) \\ 2g(x) &= 0 \\ g(x) &= 0 \end{align*}

  6. You remember learning in a physics class that when you throw a ball in the air, its height (in meters) as a function of time (in seconds) \(h(t)\) can be modeled by a quadratic. Unfortunately, you completely forget what this quadratic is. As an experiment, you throw a ball in the air. It travels to a maximum height of 4.9 meters and lands back in your hand after 2 seconds. Use this information to write an expression for \(h(t)\).

    Solution

    From the problem, we know the ball is at height 0 when \(t = 0\) and \(t = 2\), so \(0\) and \(2\) are booth roots of \(h\). Since \(h\) is quadratic and we know both its roots, we know that \[ h(t) = a(t - 0)(t - 2) = at(t - 2) \] for some constant \(a\). We also know that a quadratic attains its maximum exactly between its two roots, meaning at \(t = 1\). Thus we know \(h(1) = 4.9\). We can use this to solve for \(a\): \[ h(1) = a\cdot 1(1 - 2) = -a = 4.9 \implies a = -4.9 \] Thus \(h(t) = -4.9t(t - 2)\).

  7. European paper comes in “A” sizes: A0, A1, A2, etc. A0 paper is huge and is used for posters, while A4 paper is close to “normal” US Letter paper. When you cut a peice of A0 paper in half, you get two A1 sheets. When you cut an A1 sheet, you get two A2 sheets, and so on. At the same time, the ratio of width to height \(w/h\) remains the same each time you cut it in half. What must this ratio be? (Hint: The tricky part here is setting things up. Try drawing a picture. At the end of the day you will be solving a quadratic.)

    Solution

    Let \(w_{A0}\) and \(h_{A0}\) be the width and height of a peice of A0 paper. A peice of A1 paper is made from cutting the height in half and rotating 90 degrees, so that the width and height are interchanged. Thus the width and height of A1 paper are given by \(w_{A1} = h_{A0}/2\) and \(h_{A1} = w_{A0}\). We know this ratios are the same, meaning

    \begin{align*} \frac{w_{A0}}{h_{A0}} &= \frac{w_{A1}}{h_{A1}} \\ \frac{w_{A0}}{h_{A0}} &= \frac{h_{A0}/2}{w_{A0}} && \text{substituting for $w_{A1}$ and $h_{A1}$} \\ \left(\frac{w_{A0}}{h_{A0}}\right)^2 &= \frac{1}{2} && \text{multiplying both sides by $\frac{w_{A0}}{h_{A0}}$}\\ \left|\frac{w_{A0}}{h_{A0}}\right| &= \frac{1}{\sqrt 2} && \text{taking the square root of both sides} \end{align*}

    Since width and height are always positive numbers, we know \(w_{A0}/h_{A0} = 1/\sqrt{2}\).

  8. In music, the note A4 has a frequency of 440 Hz. The smallest gap between notes is called a half-step, and 12 half-steps make an octave. Every time you move up an octave, the frequency of a note doubles. Write an function \(f(s)\), giving the frequency (in Hz) of a note as a function of the number of half-steps \(s\) above A4. The note C4 (often called middle C) is 9 half-steps below A4. What is its frequency? (Hint: First think about frequency as a function of octaves above A4, and then octaves as a function of half-steps, and then put them together.)

    Solution

    To see the pattern, try a few examples:

    Octaves above A4 Frequency (in Hz)
    \(0\) \(440\)
    \(1\) \(440\cdot 2 = 880\)
    \(2\) \(440\cdot 2\cdot 2 = 1760\)
    \(3\) \(440\cdot 2\cdot 2\cdot 2 = 3520\)

    So it looks like the frequency is \(440 \cdot 2^o\), where \(o\) is the number of octaves. In terms of half-steps, we have \(o = s/12\), so \[ f(s) = 440 \cdot 2^{s/12} \] The frequency of middle C is then given by \[ f(-9) = 440 \cdot 2^{-9/12} \approx 261.626 \] so the answer is 261.626 Hz.

  9. In many disciplines, it can be useful to plot data or a function on a grid where the axes don’t represent the values \(x\) and \(y\) directly, but rather \(\log_{10} x\) and \(\log_{10} y\). In this exercise, we will explore what this transformation does to a few familiar functions. Let \(u\) and \(v\) be the points on this grid, so \(u = \log_{10}x\) and \(v = \log_{10} y\). If we want to go the other way, we have \(x = 10^u\) and \(y = 10^v\).

    To see what \(y = 5x\) looks like on a log-log plot, re-write it in terms of \(u\) and \(v\), and write \(v\) as a function of \(u\).

    What does \(y = 1/x\) look like on this plot?

    A function that looks linear on a log-log plot is of the form \(v = mu + b\), or equivalently, \(\log_{10} y = m\log_{10} x + b\). To see what function this represents, write this as a function of \(y\) and simplify.

    If a function looks like \(v = u^2\) on a log-log plot, what was the original function \(y\) in terms of \(x\)?

    Solution

    For the function \(y = 5x\), we have

    \begin{align*} 10^v &= 5 \cdot 10^u \\ \log_{10}\left(10^v\right) &= \log_{10}\left(5 \cdot 10^u\right) \\ v &= u + \log_{10} 5 \end{align*}

    So the graph is linear, with slope 1 and intercept \(\log_{10}5\).

    For the function \(y = 1/x\), we have

    \begin{align*} 10^v &= \frac{1}{10^u} = 10^{-u} \\ \log_{10}(10^v) &= \log_{10}(10^{-u}) \\ v &= -u \end{align*}

    So the graph is again linear, with slope \(-1\) and intercept 0.

    A function which looks linear on this log-log plot has \(v = mu + b\). Substituting \(v = \log_{10} y\) and \(u = \log_{10} x\) gives

    \begin{align*} \log_{10} y &= m\log_{10}x + b \\ 10^{\log_{10} y} &= 10^{m \log_{10} x + b} \\ y &= \left(10^{\log_{10} x}\right)^m \cdot 10^b \\ &= 10^b \cdot x^m \end{align*}

    So when the log-log plot is linear, \(y\) is some power of \(x\), times a constant multiple. The power is determined by the slope \(m\), and the multiple is determined by the intercept \(b\). This is consistent with the first two examples we saw.

    If a function looks like \(v = u^2\) on a log-log plot, we have

    \begin{align*} \log_{10} y &= (\log_{10} x)^2 \\ 10^{\log_{10} y} &= 10^{(\log_{10} x)^2} = \left(10^{\log 10_x}\right)^{\log_{10} x} \\ y &= x^{\log_{10} x} \end{align*}

    Note that there may be more than one good way to write this function. Comparing this example with the previous ones, it should make sense that this function grows faster than any fixed power of \(x\).