Week 4 Worksheet

  1. On what set is the function \(\displaystyle g(x) = \frac{x - 2}{x^2 - 2}\) continuous?

    Solution

    This is a rational function, so it is continuous everywhere in its domain. It is discontinuous at \(x = \pm\sqrt{2}\). Taking the limit at these points,

    \begin{align*} \lim_{x\to\sqrt 2^+} \frac{x - 2}{(x - \sqrt 2) (x + \sqrt 2)} &= -\infty & \lim_{x\to\sqrt 2^-} \frac{x - 2}{(x - \sqrt 2) (x + \sqrt 2)} &= +\infty \\ \lim_{x \to -\sqrt 2^+} \frac{x - 2}{(x - \sqrt 2)(x + \sqrt 2)} &= +\infty & \lim_{x\to -\sqrt 2^-} \frac{x - 2}{(x - \sqrt 2)(x + \sqrt 2)} &= -\infty \end{align*}

  2. Find the set where \(f\) is continuous, where

    \[f(x) = \begin{cases} x + 1 &\quad x < -1 \\ x^2 - x &\quad -1 \leq x < 1 \\ 2 &\quad x = 1 \\ \frac{-x^2 + 3x - 2}{x - 2} &\quad x > 1 . \end{cases}\]

    Solution

    First, \(2\) is not in the domain, so we have a discontinuity at 2.

    Since this is a piecewise function which changes at \(1\) and \(-1\), we need to check those points for continuity. To do this, we take the limits and compare with values

    \begin{align*} \lim_{x \to -1^-} f(x) &= \lim_{x \to -1^-} x + 1 = 0 & \lim_{x \to -1^+} f(x) &= \lim_{x \to -1^+} x^2 - x = 2 & f(-1) &= 2 \\ \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} x^2 - x = 0 & \lim_{x \to -1^+} f(x) &= \lim_{x \to -1^+} \frac{-x^2 + 3x - 2}{x - 2} = 0 & f(-1) &= 0 \end{align*}

    At \(x = -1\), the limits from either side do not agree, so the limit does not exist and \(f\) is discontinuous. At \(x = 1\), the limit exists and agrees with the original function, so \(f\) is continuous.

  3. Find the set where \(h\) is continuous, where

    \[ h(x) = \begin{cases} \sin\left(\frac{1}{x}\right) &\quad x \neq 0\\ 0 &\quad x = 0. \end{cases} \]

    Solution

    This funciton is defined everywhere, and at \(x \neq 0\) it is given by \(\sin(1/x)\), which is continuous. Thus the only place it might not be continuous is at \(x = 0\). Now, we take the limit from either side:

    \[ \lim_{x\to 0^-} h(x) = \lim_{x \to 0^-} \sin\left(\frac{1}{x}\right) = \lim_{y \to -\infty} \sin y \] \[ \lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \sin\left(\frac{1}{x}\right) = \lim_{y \to \infty} \sin y \]

    Now, as \(y\) approaches \(\pm\infty\), the function \(\sin y\) oscillates, without approaching any specific value, so the limit does not exist. Thus \(h\) is not continuous at \(x = 0\).

  4. What is the average rate of change of the function \(f(x) = 1 - x^2\) on the interval \([-1, 5]\)?

    Solution

    \[ \frac{f(5) - f(-1)}{5 - (-1)} = \frac{(1 - 5^2) - (1 - (-1)^2)}{6} = \frac{-24}{6} = -4 \]

  5. Find the derivative of \(f(x) = x - 3x^2\) at the point \(x = 4\).

    Solution \begin{align*} f'(4) = \lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} &= \lim_{h \to 0} \frac{(4 + h - 3(4 + h)^2) - (4 - 3(4)^2)}{h} \\ &= \lim_{h \to 0} \frac{-23h - 3h^2}{h} \\ &= \lim_{h \to 0} -23 - h = -23 \end{align*}

  6. Write the equation of a line tangent to the curve \(y = x^3 - x\) at the point \((1,0)\).

    Solution

    The slope of the tangent line is \[ f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{(1 + h)^3 - (1 + h)}{h} = \lim_{h \to 0} \frac{2h + 3h^2 + h^3}{h} = \lim_{h \to 0} 2 + 3h + h^2 = 2 \]

    The line with slope passing through the point \((1,0)\) is \(y = 2(x - 1)\).

  7. What is the derivative of \(f(x) = |x|\)? (Hint: It’s not defined everywhere.)

    Solution

    We have to consider a few cases. If \(x\) is positive, then for any sufficiently small \(h\), \(|x + h|\) will also be positive. Thus \(|x| = x\) and \(|x + h| = x + h\). Therefore,

    \[ f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h} = \lim_{h \to 0} \frac{x + h - x}{h} = 1 \]

    If \(x\) is negative, then for any sufficiently small \(h\), \(|x + h|\) will also be negative, and so \(|x + h| = - x - h\) and \(|x| = -x\). Thus

    \[ f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h} = \lim_{h \to 0} \frac{-x-h - (-x)}{h} = -1 \]

    Last, we consider when \(x = 0\). In this case,

    \[ f'(0) = \lim_{h \to 0} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0} \frac{|h|}{h} \]

    Last week, we showed this limit does not exist. Therefore \(f\) is not differentiable at \(x = 0\). To summarize,

    \[ f'(x) = \begin{cases} 1 &\quad x > 0 \\ -1 &\quad x < 0 \\ \text{undefined} &\quad x = 0. \end{cases} \]

  8. According to your textbook, the cost in dollars to produce \(x\) tacos is \(C(x) = -\frac{3}{800}x^2 + \frac{3}{2}x + 1000\), for \(0 \leq x \leq 180\). Find a formula for marginal cost at a level of \(x\) tacos. What is the marginal cost at a level of 100 tacos?

    Solution

    Recall that marginal cost is the derivative of the cost function.

    \begin{align*} C'(x) &= \lim_{h \to 0} \frac{-\frac{3}{800}(x+h)^2 + \frac{3}{2}(x+h) + 1000 -(-\frac{3}{800}x^2 + \frac{3}{2}x + 1000)}{h} \\ &= \lim_{h \to 0} \frac{-\frac{3}{800}h^2 - \frac{3}{400}xh + \frac{3}{2}h}{h} \\ &= \lim_{h \to 0} -\frac{3}{800}h - \frac{3}{400}x + \frac{3}{2} = -\frac{3}{400}x + \frac{3}{2} \end{align*}

    In particular, \(f'(100) = -\frac{3}{400}(100) + \frac{3}{2} = \frac{3}{4} = 0.75\) dollars per taco.

  9. Find the derivative of \(f(x) = \sqrt[3]{x}\). (Hint: For square roots, you multiply by the conjugate and take advantage of the fact that \((a-b)(a+b) = a^2-b^2\). Try to do something analogous using \((a-b)(a^2 + ab + b^2) = a^3-b^3\).)

    Solution

    Write \(f(x) = x^{1/3}\).

    \begin{align*} f'(x) &= \lim_{h \to 0} \frac{(x + h)^{1/3} - x^{1/3}}{h} \\ &= \lim_{h \to 0} \frac{((x + h)^{1/3} - x^{1/3})((x + h)^{2/3} + x^{2/3} + x^{1/3}(x + h)^{1/3})}{h((x + h)^{2/3} + x^{2/3} + x^{1/3}(x + h)^{1/3})} \\ &= \lim_{h \to 0} \frac{(x + h) - x}{h((x + h)^{2/3} + x^{2/3} + x^{1/3}(x + h)^{1/3})} \\ &= \lim_{h \to 0} \frac{1}{(x + h)^{2/3} + x^{2/3} + x^{1/3}(x + h)^{1/3}} \\ &= \frac{1}{x^{2/3} + x^{2/3} + x^{2/3}} = \frac{1}{3x^{2/3}} \end{align*}

    It may be enlightening to realize that this is implicitly using the formula

    \[ (a - b)(a^2 + b^2 + ab) = a^3 - b^3. \]

    This is the analogue of “multiply by the conjugate” for cubes. (In general, we have

    \[ (a - b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1}) = a^n - b^n \]

    which you can use to calculate the derivative of \(\sqrt[n]{x}\) from first principals).

    1. Expand out \((x+h)^2\), \((x + h)^3\), and \((x + h)^4\).

    2. You may already notice some patterns, but the only pattern we need for now is

      \[(x + h)^n = x^n + nx^{n-1}h + (\text{terms divisible by $h^2$})\]

      Argue for why this equation must hold for all \(n\). (Hint: There are many ways to go here, but one way is to count how many ways you can pick a term from each factor in \((x + h)(x+h)\cdots(x+h)\), so that the product comes out to \(x^{n-1}h\).)

    3. Use this to calculate the derivative of \(f(x) = x^n\). (You may already know this rule from a previous class, but I want you to show why this rule works.)