Week 9 Worksheet

Decide if each of the following is true or false. Justify your answer.

In what follows, \(f\) is a function, and \(a,b,c\) are constants.

1. If \(f'(x) > 0\) on an interval, then \(f\) is positive on that interval.

   Solution

   False. To find a counterexample, we need a function which is increasing and negative. \(f(x) = x\) on the interval \((-2,-1)\) has \(f'(x) = 1 > 0\) everywhere, and is negative on the given interval.

2. If \(f'(c)\) exists, then \(f''(c)\) exists.

   Solution

   False. The most straightforward way to find a counterexample is to find a function such that the graph of \(f'\) has a sharp corner. If we could find an \(f\) such that \(f'(x) = |x|\), then we would be done, since then \(f'(0)\) would exist but \(f''(0)\) wouldn’t.

   Recall that we can write \(|x|\) as a peicewise function:

   \[ |x| = \begin{cases} x &\quad x \geq 0 \\ -x &\quad x < 0 \end{cases} \]

   We want this to be \(f'\), so our guess for \(f\) is \[ f(x) = \begin{cases} \frac{1}{2}x^2 &\quad x \geq 0 \\ -\frac{1}{2}x^2 &\quad x < 0 \end{cases} \]

   It’s clear that \(f'(x) = |x|\) for \(x \neq 0\), using the power rule. (This is by the principal that the limit/derivative of a piecewise function away from a jump is just the limit/derivative of that “piece”). At \(x=0\), the limit of the difference quotient will involve both parts of the peicewise function, so we need to use the definition of the derivative and take the limit from either side:

   \[ \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{1}{2}h^2}{h} = \lim_{h \to 0^+} \frac{1}{2} h = 0 \] \[ \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-\frac{1}{2}h^2}{h} = \lim_{h \to 0^-} -\frac{1}{2} h = 0 \]

   The limit from either side is 0, so \(f'(0) = 0 = |0|\), as desired. We did all this work to show that \(f'(x) = |x|\) for all \(x\), whether \(x=0\) or \(x\neq 0\). By the remarks at the beginning, this disproves the statement.

3. If \(f'(c) = 0\) and \(f''(c) = 0\), then \(c\) is neither a min nor max of \(f\).

   Solution

   False. As a counterexample, take \(f(x) = x^4\). Then \(f'(0) = f''(0) = 0\), but \(0\) is a local (and in fact global) minimum of \(x^4\), since \(x^4 \geq 0\) for all \(x\).

4. If \(f\) is an odd function whose domain contains 0, then \(f(0) = 0\).

   Solution

   True. Here is a proof: Suppose \(f\) is an odd function. By definition, this means \(f(-x) = -f(x)\) for all \(x\) in the domain of \(f\). In particular, this means \(f(0) = f(-0) = -f(0)\). Re-arranging, we get \(2f(0) = 0\), so \(f(0) = 0\).

5. It’s possible for a function \(f\) defined on \((a,b)\) to not have any relative extrema.

   Solution

   True. We want to show this is possible, which means we just need to find one example. Take \(f(x) = x\) on the interval \((-1,1)\). A relative extremum must occur at a critical point or an end point. But \((-1, 1)\) has no endpoints, and \(f(x) = x\) has no critical points, so there cannot be any relative extrema.

1. \(f(x) = -x^2 + 4x - 8\).

   Solution

   The domain of \(f\) is all real numbers, so there are no end points. Thus the only place a local extremum can occur is at a critical point.

   We have \(f'(x) = -2x + 4\). This is defined everywhere. Solving for \(f'(x) = 0\) gives \(x = 2\), so \(f\) has a single critical point at \(2\). Using the second derivative test, \(f''(x) = -2\), so \(f\) has a local maximum at \(2\). Its value is \(f(2) = -(2)^2 + 4(2) - 8 = -4\).

2. \(g(x) = \frac{xe^x}{x - 1}\).

   Solution

   The domain of \(f\) is \((-\infty,1) \cup (1,\infty)\), which again has no end points, so we only have to look at critical points.

   We have

   \[ g'(x) = \frac{(xe^x + e^x)(x - 1) - xe^x}{(x - 1)^2} = \frac{e^x(x^2 - x - 1)}{(x - 1)^2}. \]

   This is defined everywhere but \(x = 1\), but this point is not in the domain of \(g\) so we can ignore it. This is zero exactly when \(x^2 - x - 1 = 0\). Using the quadratic formula, we know this is at \(x = \frac{1 \pm \sqrt{5}}{2}\). To determine if each of these is a min or max, use the first derivative test, meaning look at the sign of \(g'(x)\). Since \(e^x\) and \((x - 1)^2\) are positive on the domain of \(g\), the sign of \(g'(x)\) is the same as the sign of \(x^2 - x - 1\). This is a quadratic where the leading coefficient is positive, so we know its graph looks like “\(\cup\)”, as opposed to “\(\cap\)”. Thus \(g'\) is positive on \((-\infty, \frac{1-\sqrt{5}}{2}) \cup (\frac{1 + \sqrt{5}}{2},\infty)\), and \(g'\) is negative on \((\frac{1 - \sqrt{5}}{2},\frac{1+\sqrt{5}}{2})\). Thus \(g\) has a local max at \(\frac{1 - \sqrt{5}}{2}\), and a local min at \(\frac{1 + \sqrt{5}}{2}\). The values at these points are

   \[ g\left(\frac{1 - \sqrt{5}}{2}\right) = \frac{\left(\frac{1 - \sqrt{5}}{2}\right)e^{\left(\frac{1 - \sqrt{5}}{2}\right)}}{\left(\frac{1 - \sqrt{5}}{2}\right) - 1} \qquad \qquad g\left(\frac{1 + \sqrt{5}}{2}\right) = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)e^{\left(\frac{1 + \sqrt{5}}{2}\right)}}{\left(\frac{1 + \sqrt{5}}{2}\right) - 1} \]

3. \(h(x) = \sqrt[3]{x - 2} + 7x\).

   Solution

   The domain of \(h\) is all real numbers, so again there are no end points to check. We will look for critical points.

   We have \(h'(x) = \frac{1}{3 \sqrt[3]{(x - 2)^2}} + 7\). This is undefined at \(x = 2\), so \(h\) has a critical point at 2. Solving for when \(h'(x) = 0\), we get

   \[ \frac{1}{\sqrt[3]{(x - 2)^2}} = -21. \]

   We know \((x - 2)^2\) is non-negative, so in fact the left side of the equation must be non-negative. This is impossible, which means there are no \(x\) such that \(h'(x) = 0\).

   Thus the only critical point is at \(2\). We will use the first derivative test. For the same reason as before, \(\frac{1}{3\sqrt[3]{(x - 2)^2}}\) is positive wherever it is defined, which means that \(h'(x)\) is positive everywhere it is defined. Thus \(h(x)\) is increasing both to the left and right of \(2\), so this is neither a min nor a max. Thus \(h'\) has no local extrema.

4. \(k(x) = 10 - x^2\), defined on \((-5, 6]\).

   Solution

   The domain is \((-5, 6]\), so we do have an endpoint at \(x = 6\) that we will need to check. The derivative is \(k'(x) = -2x\), so \(k'(0) = 0\), meaning \(0\) is a critical point. The derivative is positive on \((-5, 0)\) and negative on \((0, 6]\), meaning that \(k\) has a local max at 0 and a local min at 6. The values here are

   \[ k(0) = 10 \qquad \qquad k(6) = 10 - 6^2 = -26 \]

1. \(\ell(x) = x^2e^{2x}\)
2. \(m(x) = \frac{-4x}{1 + x}\)

In probability, one of the most commonly occuring distributions is the normal distribution, or “bell curve.” It depends on two constants: the mean \(\mu\), and the standard deviation \(\sigma\). The formula for the distibution is given by

\[ f(x) = \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2} \]

This function has a single local maximum. Where is it, and what is the value of \(f\) there? Where are the points of inflection?