Coordinated Calculus

When is \(f'(x)=0\text{?}\) What else do you need to look for?

\(-1\) is a critical number that is not a relative extremum. \(3\) is a local maximum.

Since we already have \(f'(x)\) written in factored form, it is straightforward to find the critical numbers of \(f\text{.}\) Because \(f'(x)\) is defined for all values of \(x\text{,}\) we need only determine where \(f'(x) = 0\text{.}\) From the equation

and the zero product property³The zero product property says that if \(ab=0\text{,}\) where \(a\) and \(b\) are expressions representing real numbers (e.g. \(a=(3-x)\text{,}\) where \(x\) is a real number), then it must be the case that \(a=0\) or \(b=0\) (or both)., it follows that \(x = 3\) and \(x = -1\) are critical numbers of \(f\text{.}\) (There is no value of \(x\) that makes \(e^{-2x} = 0\text{.}\))

Next, to apply the first derivative test, we'd like to know the sign of \(f'(x)\) at inputs near the critical numbers. Because the critical numbers are the only locations at which \(f'\) can change sign, it follows that the sign of the derivative is the same on each of the intervals created by the critical numbers: for instance, the sign of \(f'\) must be the same for every \(x \lt -1\text{.}\) What this means is that we can choose (carefully) where to evaluate the derivative in order to ascertain its sign on a given interval. Since \(f'(-1.000001)\) and \(f'(-2)\) must have the same sign, we may as well evaluate \(f'(x)\) at \(-2\) to figure out the sign of \(f'\) for \(x\lt -1\text{.}\) We create a first derivative sign chart (displayed below, with explanation following) to summarize the sign of \(f'\) on the relevant intervals, along with the corresponding behavior of \(f\text{.}\)

To produce the first derivative sign chart in Figure3.6, we start by marking the critical numbers \(-1\) and \(3\) on the number line. We then identify the sign of each factor of \(f'(x)\) at one selected point in each interval. The intervals in this example are \(x\lt-1\text{,}\) \(-1\lt x\lt3\text{,}\) and \(x\gt3\text{;}\) we will choose \(x=-2\text{,}\) \(x=0\text{,}\) and \(x=5\) for our selected points. The process with \(x=-2\) is laid out below:

For \(x \lt -1\text{,}\) we use the value \(x=-2\) to determine the sign of \(e^{-2x}\text{,}\) \((3-x)\text{,}\) and \((x+1)^2\text{.}\) We note that \(e^{-2x}\) is positive regardless of the value of \(x\text{,}\) \((x+1)^2\) is positive whenever it is nonzero which is everywhere within the intervals of interest since we intentionally plucked out the zeros and that \((3-x)\) is also positive at \(x = -2\text{.}\) Hence each of the three terms in \(f'\) is positive, and we indicate this by writing \(+++\) above the interval \(x\lt-1\text{.}\) Taking the product of three positive terms results in a positive value for \(f'\text{,}\) so we denote the sign of \(f'\) by a \(+\) above the appropriate interval in the chart. Finally, since \(f'\) is positive on that interval, we know that \(f\) is increasing, so we write INC to represent the behavior of \(f\text{.}\)

In a similar fashion, we find that \(f'\) is positive (because \(3-(0)\gt0\) and the other terms are also still positive) and \(f\) is increasing on \(-1 \lt x \lt 3\text{,}\) and that \(f'\) is negative (because \(3-(5)\lt0\) but the other terms are still positive) and \(f\) is decreasing for \(x \gt 3\text{.}\)

Now we look for critical numbers at which \(f'\) changes sign. In this example, \(f'\) changes sign only at \(x = 3\text{,}\) from positive to negative, so \(f\) has a relative maximum at \(x = 3\text{.}\) Although \(f\) has a critical number at \(x = -1\text{,}\) since \(f\) is increasing both before and after \(x = -1\text{,}\) \(f\) has neither a minimum nor a maximum at \(x = -1\text{.}\)

1. For which \(x\) is \(g'(x) = 0\text{?}\)

2. Note that \((x-1)^2\) is positive for all \(x \ne 1\text{.}\)

3. Use your first derivative sign chart from (b).

4. Try expanding the numerator before evaluating the limit.

5. Think carefully about the information from the sign chart found in (b). Your answers to part (c) and (d) should also support your graph.

1. \(x = -4\) or \(x = 1\text{.}\)

2. \(g\) has a local maximum at \(x = -4\) and neither a max nor min at \(x = 1\text{.}\)

3. \(g\) does not have a global minimum; it is unclear (at this point in our work) if \(g\) increases without bound, so we can't say for certain whether or not \(g\) has a global maximum.

4. \(\lim_{x \to \infty} g'(x) = \infty\text{.}\)

5. A possible graph of \(g\) is the following.

   

1. Since \(g'(x) = \frac{(x+4)(x-1)^2}{x-2}\text{,}\) we see that \(g'(x) = 0\) implies that \(x = -4\) or \(x = 1\text{.}\) Although \(x = 2\) makes \(g'\) undefined, we are told that \(g\) has a vertical asymptote at \(x = 2\text{.}\) So \(x = 2\) is not in the domain of \(g\text{,}\) and hence is technically not a critical number of \(g\text{.}\) Nonetheless, we place \(x = 2\) on our first derivative sign chart since the vertical asymptote is a location at which \(g'\) may change sign.

2. The first derivative sign chart shows that:

   • \(g'(x) \gt 0\) for \(x \lt -4\text{,}\)
   • \(g'(x) \lt 0\) for \(-4 \lt x \lt 1\text{,}\)
   • \(g'(x) \lt 0\) for \(1 \lt x \lt 2\text{,}\) and
   • \(g'(x) \gt 0\) for \(x \gt 2\text{.}\)

   By the first derivative test, \(g\) has a local maximum at \(x = -4\) and neither a max nor min at \(x = 1\text{.}\) As these are the only two critical numbers, these are the only two locations for possible extrema. (Note: although \(g\) changes from decreasing to increasing at \(x = 2\text{,}\) this is due to a vertical asymptote, and \(g\) does not have a minimum there.)

3. Because \(g\) is decreasing as \(x \to 2^-\) (where \(g\) has a vertical asymptote), \(g\) does not have a global minimum. For \(x \gt 2\text{,}\) \(g\) is always increasing, which suggests that \(g\) does not have a global maximum (though we do not know for sure that \(g\) increases without bound; this is explored in part (d)).

4. We observe that

   \begin{align*} \lim_{x \to \infty} g'(x) =\mathstrut \amp \lim_{x \to \infty} \frac{(x+4)(x-1)^2}{x-2}\\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^3 + 2x^2 - 7x + 4}{x-2} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}\\ =\mathstrut \amp \lim_{x \to \infty} \frac{x^2 + 2x - 7 + \frac{4}{x}}{1 - \frac{2}{x}}\\ =\mathstrut \amp \infty \end{align*}

   Since \(g'(x) \to \infty\) as \(x \to \infty\text{,}\) this tells us that \(g\) increases without bound as \(x \to \infty\text{.}\)

5. From all of our work above, we know that \(g\) has a local maximum at \(x = -4\text{,}\) a horizontal tangent line with neither a max nor min at \(x = 1\text{,}\) and a vertical asymptote at \(x = 2\text{,}\) plus \(g\) and \(g'\) both increase without bound as \(x \to \infty\text{.}\) Thus, a possible graph of \(y=g(x)\) is the following.

   

Since we know \(f'(x) = 3x^4 - 9x^2\text{,}\) we can find the critical numbers of \(f\) by solving \(3x^4 - 9x^2 = 0\text{.}\) Factoring, we observe that

so that \(x = 0, \pm\sqrt{3}\) are the three critical numbers of \(f\text{.}\) The first derivative sign chart for \(f\) is given below in Figure3.10.

We see that \(f\) is increasing on the intervals \((-\infty, -\sqrt{3})\) and \((\sqrt{3}, \infty)\text{,}\) and \(f\) is decreasing on \((-\sqrt{3},0)\) and \((0, \sqrt{3})\text{.}\) By the first derivative test, this information tells us that \(f\) has a local maximum at \(x = -\sqrt{3}\) and a local minimum at \(x = \sqrt{3}\text{.}\) Although \(f\) also has a critical number at \(x = 0\text{,}\) neither a maximum nor minimum occurs there since \(f'\) does not change sign at \(x = 0\text{.}\)

Next, we move on to investigate concavity. Differentiating \(f'(x) = 3x^4 - 9x^2\text{,}\) we see that \(f''(x) = 12x^3 - 18x\text{.}\) Since we are interested in knowing the intervals on which \(f''\) is positive and negative, we first find where \(f''(x) = 0\text{.}\) Observe that

This equation has solutions \(x = 0, \pm\sqrt{\frac{3}{2}}\text{.}\) Building a sign chart for \(f''\) in the exact same way we do for \(f'\text{,}\) we see the result shown below in Figure3.11.

Therefore \(f\) is concave down on the intervals \(\left(-\infty, -\sqrt{\frac{3}{2}}\right)\) and \(\left(0, \sqrt{\frac{3}{2}}\right)\text{,}\) and concave up on \(\left(-\sqrt{\frac{3}{2}},0\right)\) and \(\left(\sqrt{\frac{3}{2}}, \infty\right)\text{.}\)

Putting all of this information together, we now see a complete and accurate possible graph of \(f\) in Figure3.12.

The point \(A = (-\sqrt{3}, f(-\sqrt{3}))\) is a local maximum, because \(f\) is increasing prior to \(A\) and decreasing after; similarly, the point \(E = (\sqrt{3}, f(\sqrt{3})\) is a local minimum. Note, too, that \(f\) is concave down at \(A\) and concave up at \(E\text{,}\) which is consistent both with our second derivative sign chart and the second derivative test. At points \(B\) and \(D\text{,}\) concavity changes, as we saw in the results of the second derivative sign chart in Figure3.11. Finally, at point \(C\text{,}\) \(f\) has a critical point with a horizontal tangent line, but neither a maximum nor a minimum occurs there, since \(f\) is decreasing both before and after \(C\text{.}\) It is also the case that concavity changes at \(C\text{.}\)

While we completely understand where \(f\) is increasing and decreasing, where \(f\) is concave up and concave down, and where \(f\) has relative extrema, we do not know any specific information about the \(y\)-coordinates of points on the curve. For instance, while we know that \(f\) has a local maximum at \(x = -\sqrt{3}\text{,}\) we don't know the value of that maximum because we do not know \(f(-\sqrt{3})\text{.}\) Any vertical translation of our sketch of \(f\) in Figure3.12 would satisfy the given criteria for \(f\text{.}\)

1. What must be true of \(g''(x)\) at a point of inflection?

2. Use the given graph to decide where \(g''\) is positive and negative.

3. What does the second derivative test say?

4. Can you guess a formula for \(g''(x)\) based on its graph?

1. \(x = -1\) is an inflection point of \(g\text{.}\)

2. \(g\) is concave up for \(x \lt -1\text{,}\) concave down for \(-1 \lt x \lt 2\text{,}\) and concave down for \(x \gt 2\text{.}\)

3. \(g\) has a local minimum at \(x = -1.67857351\text{.}\)

4. \(g\) is a degree 5 polynomial.

1. Based on the given graph of \(g''\text{,}\) the only point at which \(g''\) changes sign is \(x = -1\text{,}\) so this is the only inflection point of \(g\text{.}\)

2. Note that \(g''(x) \gt 0\) for \(x \lt -1\text{,}\) \(g''(x) \lt 0\) for \(-1 \lt x \lt 2\text{,}\) and \(g''(x) \lt 0\) for \(x \gt 2\text{.}\) This tells us that \(g\) is concave up for \(x \lt -1\text{,}\) concave down for \(-1 \lt x \lt 2\text{,}\) and concave down for \(x \gt 2\text{.}\)

3. Given that \(g'(-1.67857351) = 0\text{,}\) we know that \(g\) has a horizontal tangent line at this critical number. Additionally, the graph of \(y=g''(x)\) shows that \(g''( -1.67857351) \gt 0\text{,}\) so we infer that \(g\) is concave up at \(x\)-values near \(-1.67857351\text{.}\) The second derivative test allows us to conclude that \(g\) has a local minimum at \(x = -1.67857351\text{.}\)

4. As seen in the given graph, since \(g''\) has a simple zero at \(x = -1\) and a repeated zero at \(x = 2\text{,}\) it appears that \(g''\) is a degree 3 polynomial. If so, then \(g'\) is a degree 4 polynomial, and \(g\) is a degree 5 polynomial.

1. Be sure to try some values between 1.5 and 2.

2. Treat \(k\) as an arbitrary constant in computing \(h'(x)\) and \(h''(x)\text{.}\)

3. What is the largest value of \(k\sin(kx)\text{?}\) How many times can \(y = k\sin(kx)\) intersect the line \(y = 2x\text{?}\)

1. In the graph below, \(h(x) = x^2 + \cos(3x)\) is given in dark blue, while \(h(x) = x^2 + \cos(1.6x)\) is shown in light blue.

   

2. If \(\frac{2}{k^2} \gt 1\text{,}\) then the equation \(\cos(kx) = \frac{2}{k^2}\) has no solution. Hence, whenever \(k^2 \lt 2\text{,}\) or \(k \lt \sqrt{2} \approx 1.414\text{,}\) it follows that the equation \(\cos(kx) = \frac{2}{k^2}\) has no solutions \(x\text{,}\) which means that \(h''(x)\) is never zero (indeed, for these \(k\)-values, \(h''(x)\) is always positive so that \(h\) is always concave up). On the other hand, if \(k \ge \sqrt{2}\) then \(\frac{2}{k^2} \le 1\text{,}\) which guarantees that \(\cos(kx) = \frac{2}{k^2}\) has infinitely many solutions due to the periodicity of the cosine function. At each such point, \(h''(x)\) changes sign if and only if \(k\gt\sqrt{2}\text{.}\) (In the case where \(k=\sqrt2\text{,}\) the function \(h''\) has infinitely many zeros but is always non-negative.) Therefore \(h\) has infinitely many inflection points whenever \(k \gt \sqrt{2}\text{,}\) and no inflection points otherwise.

3. To see why \(h\) can only have a finite number of critical numbers regardless of the value of \(k\text{,}\) consider the equation

   \begin{equation*} 0 = h'(x)\text{,} \end{equation*}

   which implies that \(2x = k\sin(kx)\text{.}\) Since \(-1 \le \sin(kx) \le 1\text{,}\) we know that \(-k \le k\sin(kx) \le k\text{.}\) Once \(|x|\) is sufficiently large, we are guaranteed that \(|2x| \gt k\text{,}\) which means that for large values of \(x\text{,}\) the graphs of \(y=2x\) and \(y=k\sin(kx)\) cannot intersect. Moreover, for relatively small values of \(x\text{,}\) the functions \(2x\) and \(k\sin(kx)\) can only intersect finitely many times since \(k\sin(kx)\) only oscillates a finite number of times on a fixed interval. This is why \(h\) can only have a finite number of critical numbers, regardless of the value of \(k\text{.}\)

1. In the graph below, \(h(x) = x^2 + \cos(3x)\) is given in dark blue, while \(h(x) = x^2 + \cos(1.6x)\) is shown in light blue.

   

   Close inspection of the light blue graph reveals some subtle changes in concavity around \(x \approx \pm 0.5\text{.}\) For values smaller than \(1.6\text{,}\) it is very hard to visually detect any inflection points in \(h(x)\text{.}\)

2. Treating \(k\) as an arbitrary constant, we first observe that \(h'(x) = 2x - k\sin(kx)\text{.}\) Again treating \(k\) as a constant and differentiating, we find

   \begin{equation*} h''(x) = 2 - k^2\cos(kx)\text{.} \end{equation*}

   We seek the values of \(x\) at which \(h''(x) = 0\) and \(h''\) changes sign. Setting \(h''(x) = 0\) and rearranging the resulting equation, we now seek \(x\) such that

   \begin{equation*} k^2 \cos(kx) = 2\text{,} \end{equation*}

   or

   \begin{equation*} \cos(kx) = \frac{2}{k^2}\text{.} \end{equation*}

   Now, remember that \(k\) is an arbitrary positive constant and recall that \(-1 \le \cos(\theta) \le 1\) for all input values \(\theta\text{.}\) If \(\frac{2}{k^2} \gt 1\text{,}\) then the equation \(\cos(kx) = \frac{2}{k^2}\) has no solution. Hence, whenever \(k^2 \lt 2\text{,}\) or \(k \lt \sqrt{2} \approx 1.414\text{,}\) it follows that the equation \(\cos(kx) = \frac{2}{k^2}\) has no solutions, which means that \(h''(x)\) is never zero (indeed, for these \(k\)-values, \(h''(x)\) is always positive so that \(h\) is always concave up). On the other hand, if \(k \ge \sqrt{2}\) then \(\frac{2}{k^2} \le 1\text{,}\) which guarantees that \(\cos(kx) = \frac{2}{k^2}\) has infinitely many solutions due to the periodicity of the cosine function. At each such point, \(h''(x)\) changes sign if and only if \(k\gt\sqrt{2}\text{.}\) (In the case where \(k=\sqrt2\text{,}\) the function \(h''\) has infinitely many zeros but is always non-negative.) Therefore \(h\) has infinitely many inflection points whenever \(k \gt \sqrt{2}\text{,}\) and no inflection points otherwise.

3. To see why \(h\) can only have a finite number of critical numbers regardless of the value of \(k\text{,}\) recall that \(h'(x) = 2x - k\sin(kx)\) and consider the equation

   \begin{equation*} 0 = h'(x)\text{,} \end{equation*}

   which then implies that \(2x = k\sin(kx)\text{.}\) Since \(-1 \le \sin(kx) \le 1\text{,}\) we know that \(-k \le k\sin(kx) \le k\text{.}\) Once \(|x|\) is sufficiently large, we are guaranteed that \(|2x| \gt k\text{,}\) which means that for large values of \(x\text{,}\) the graphs of \(y=2x\) and \(y=k\sin(kx)\) cannot intersect. Moreover, for relatively small values of \(x\text{,}\) the functions \(2x\) and \(k\sin(kx)\) can only intersect finitely many times since \(k\sin(kx)\) only oscillates a finite number of times on a finite interval. This is why \(h\) can only have a finite number of critical numbers, regardless of the value of \(k\text{.}\)