A vertical asymptote is a place where the function becomes infinite, typically because the formula for the function has a denominator that becomes zero. For example, the reciprocal function $f(x)=1/x$ has a vertical asymptote at $x=0$, and the function $\tan x$ has a vertical asymptote at $x=\pi/2$ (and also at $x=-\pi/2$, $x=3\pi/2$, etc.). Whenever the formula for a function contains a denominator it is worth looking for a vertical asymptote by checking to see if the denominator can ever be zero, and then checking the limit at such points. Note that there is not always a vertical asymptote where the derivative is zero: $f(x)=(\sin x)/x$ has a zero denominator at $x=0$, but since $\ds \lim_{x\to 0}(\sin x)/x=1$ there is no asymptote there.

A horizontal asymptote is a horizontal line to which $f(x)$ gets closer and closer as $x$ approaches $\infty$ (or as $x$ approaches $-\infty$). For example, the reciprocal function has the $x$-axis for a horizontal asymptote. Horizontal asymptotes can be identified by computing the limits $\ds \lim_{x \to \infty}f(x)$ and $\ds \lim_{x \to -\infty}f(x)$. Since $\ds \lim_{x \to \infty}1/x=\lim_{x \to -\infty}1/x=0$, the line $y=0$ (that is, the $x$-axis) is a horizontal asymptote in both directions.

Some functions have asymptotes that are neither horizontal nor vertical, but some other line. Such asymptotes are somewhat more difficult to identify and we will ignore them.

If the domain of the function does not extend out to infinity, we should also ask what happens as $x$ approaches the boundary of the domain. For example, the function $\ds y=f(x)=1/\sqrt{r^2-x^2}$ has domain $-r < x < r$, and $y$ becomes infinite as $x$ approaches either $r$ or $-r$. In this case we might also identify this behavior because when $x=\pm r$ the denominator of the function is zero.

If there are any points where the derivative fails to exist (a cusp or corner), then we should take special note of what the function does at such a point.

Finally, it is worthwhile to notice any symmetry. A function $f(x)$ that has the same value for $-x$ as for $x$, i.e., $f(-x)=f(x)$, is called an "even function." Its graph is symmetric with respect to the $y$-axis. Some examples of even functions are: $\ds x^n$ when $n$ is an even number, $\cos x$, and $\ds \sin^2x$. On the other hand, a function that satisfies the property $f(-x)=-f(x)$ is called an "odd function." Its graph is symmetric with respect to the origin. Some examples of odd functions are: $x^n$ when $n$ is an odd number, $\sin x$, and $\tan x$. Of course, most functions are neither even nor odd, and do not have any particular symmetry.

Exercises 5.5

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts.

Ex 5.5.1 $\ds y=x^5-5x^4+5x^3$

Ex 5.5.2 $\ds y=x^3-3x^2-9x+5$

Ex 5.5.3 $\ds y=(x-1)^2(x+3)^{2/3}$

Ex 5.5.4 $\ds x^2+x^2y^2=a^2y^2$, $a>0$.

Ex 5.5.5 $\ds y=xe^x$

Ex 5.5.6 $\ds y=(e^x+e^{-x})/2$

Ex 5.5.7 $\ds y=e^{-x}\cos x$

Ex 5.5.8 $\ds y=e^x-\sin x$

Ex 5.5.9 $\ds y=e^x/x$

Ex 5.5.10 $\ds y = 4x+\sqrt{1-x}$

Ex 5.5.11 $\ds y = (x+1)/\sqrt{5x^2 + 35}$

Ex 5.5.12 $\ds y= x^5 - x$

Ex 5.5.13 $\ds y = 6x + \sin 3x$

Ex 5.5.14 $\ds y = x+ 1/x$

Ex 5.5.15 $\ds y = x^2+ 1/x$

Ex 5.5.16 $\ds y = (x+5)^{1/4}$

Ex 5.5.17 $\ds y = \tan^2 x$

Ex 5.5.18 $\ds y =\cos^2 x - \sin^2 x$

Ex 5.5.19 $\ds y = \sin^3 x$

Ex 5.5.20 $\ds y=x(x^2+1)$

Ex 5.5.21 $\ds y=x^3+6x^2 + 9x$

Ex 5.5.22 $\ds y=x/(x^2-9)$

Ex 5.5.23 $\ds y=x^2/(x^2+9)$

Ex 5.5.24 $\ds y=2\sqrt{x} - x$

Ex 5.5.25 $\ds y=3\sin(x) - \sin^3(x)$, for $x\in[0,2\pi]$

Ex 5.5.26 $\ds y=(x-1)/(x^2)$

For each of the following five functions, identify any vertical and horizontal asymptotes, and identify intervals on which the function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing.

Ex 5.5.27 $f(\theta)=\sec(\theta)$

Ex 5.5.28 $\ds f(x) = 1/(1+x^2)$

Ex 5.5.29 $\ds f(x) = (x-3)/(2x-2)$

Ex 5.5.30 $\ds f(x) = 1/(1-x^2)$

Ex 5.5.31 $\ds f(x) = 1+1/(x^2)$

Ex 5.5.32 Let $\ds f(x) = 1/(x^2-a^2)$, where $a\geq0$. Find any vertical and horizontal asymptotes and the intervals upon which the given function is concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing. Discuss how the value of $a$ affects these features.

\endinput

\begin{example}

Sketch $y=x^3-x$.

\begin{rightindent}{3.5in} First, we set $0=y'=3x^2-1$, which has solutions $x=\pm\sqrt{3}/3=\pm 0.577$. The corresponding $y$-coordinates are $-0.385$ and $+0.385$, i.e., the two "critical points" are $(0.577,-0.385)$ and $(-0.577,0.385)$. The second derivative test gives $y"=6x$, which is positive for the first point and negative for the second. Thus, the first of the two points (in the fourth quadrant) is a local minimum, and the second is a local maximum. Since $y">0$ when $x>0$ and $y" < 0$ when $x < 0$, it follows that the curve is concave upward in the right half of the graph and concave down- \end{rightindent} \hfill \begin{psfigure}{1.95in}{2.0in}{124.figND.ps} \end{psfigure} ward in the left half. The two halves are separated by the inflection point at the origin. This curve has no asymptotes. It does have symmetry, however, because it is an odd function. Its graph is shown above.

\end{example}

\begin{rightindent}{3in} \begin{example}

Sketch $y=x^3+x$.

The only difference with Example 13.1 is the $+$ in front of the $x$. But this means that the derivative $y'=3x^2+1$ is never zero, and hence there are no maxima or minima. In fact, the function is always increasing, because $y'$ is always positive. The second derivative $y"=6x$ is the same as in the last problem, and hence the concavity situation is the same. In particular, this curve also has an inflection point at the origin.

\end{example} \end{rightindent} \hfill \begin{psfigure}{2in}{2in}{124.figNE.ps} \end{psfigure}

\begin{example}

Sketch $y=x^2-\cos(2x)$ for $-\pi/2\le x\le\pi/2$.

When we set $0=y'=2x+2\sin(2x)$, we obtain the equality $\sin(2x)=-x$. However, from a quick sketch of the two curves $\sin(2x)$ and $-x$, we immediately see that the only $x$ for which they are equal is $x=0$. When $x=0$ the $y$-coordinate is $0^2-\cos(2\cdot 0)=-1$, so our critical point is $(0,-1)$. Since $y"=2+4\cos(2x)$, which is positive when $x=0$, the second derivative test tells us that $(0,1)$ is a local minimum. To find inflection points we set $0=y"=2+4\cos(2x)$. This gives $\cos(2x)=-0.5$. Looking at our table in the section on trig functions, we see that in the range from $x=0$ to $x=\pi/2$ the equality $\cos(2x)=-0.5$ holds when $2x=\frac{2}{ 3}\pi$, \begin{rightindent}{3in}

i.e., $x=\pi/3$. Since $\cos(-2x)=\cos(2x)$, the equality also holds when $x=-\pi/3$. Thus, the points $(\pi/3,1.5966)$ and $(-\pi/3,1.5966)$ are inflection points. Between these two inflection points the second derivative is positive (concave up), whereas for $x>\pi/3$ and for $x < -\pi/3$ the second derivative is negative (concave downward). Finally, note that $x^2-\cos(2x)$ is an even function, and so the graph is symmetrical with respect to the $y$-axis. \end{rightindent} \begin{psfigure}{2.0in}{1.8in}{124.figNF.ps} \end{psfigure} \end{example}

\begin{example}

Sketch $y=x^2+\frac{1}{ x}$.

Setting $0=y'=2x-\frac{1}{ x^2}$, we solve this by bringing $1/x^2$ to the left and clearing denominators: $1=2x\cdot x^2=2x^3$. So $x={\root 3\of{0.5}}= 0.7937$. The corresponding $y$-coordinate is 1.8899. Using the second derivative $y"=2+2\cdot x^{-3}$, we see that this \begin{rightindent}{3in}

is a local minimum. To find inflections, we set $2+2/x^3=0$. Clearing denominators and solving for $x$ gives $x^3=-1$, and so $x=-1$. Thus, the point $(-1,0)$ is an inflection. In this example we have an asymptote when $x=0$. To the right of the asymptote (i.e., for positive $x$), the second derivative is always positive; whereas for negative $x$ the second derivative is positive when $x < -1$ and negative when $x$ is between $-1$ and $0$. Thus, the interval $-1 < x < 0$ is a region of downward concavity; the graph is concave upward outside of this interval. Putting all this together leads to the graph at the right. \end{rightindent} \hfill \begin{psfigure}{2.0in}{2.0in}{124.figNG.ps} \end{psfigure}

There is another way to think of this example. Our function is the sum of two functions $x^2$ and $1/x$. The former function is by far the larger of the two when $x$ is large positive or large negative, whereas the reciprocal function is by far the more important when $x$ is near 0. Thus, the graph resembles $1/x$ when $x$ is near 0 and resembles $x^2$ when $x$ is far from 0. Roughly speaking, the inflection point $(-1,0)$ and the local minimum $(0.7937,1.8899)$ mark the transition from behaving like the graph of $x^2$ to behaving like the graph of $1/x$.

\end{example}

\begin{example}

Sketch (a) $y=x^3$ and (b) $y=x^4$.

(a) Setting $0=y'=3x^2$, we see that the origin is a possible maximum or minimum. However, the second derivative test tells us nothing, since $y"=6x$ also is zero when $x=0$. In fact, even though $y'=0$ when $x=0$, the origin is neither a maximum nor a minimum. Rather, it is a point of inflection, separating the concave downward region in the third quadrant from the concave upward region in the first quadrant.

(b) Again we see that both the first derivative and the second derivative vanish at the origin (and neither derivative is zero anywhere else). This time, however, the origin is a local minimum. Even though the second derivative test doesn't tell us this, we can see directly that, since $x^4$ is positive for nonzero $x$, its smallest possible value is when $x=0$. Note that the origin is not a point of inflection, even though $y"=0$ there. This is because $y"=12x^2>0$ both for $x>0$ and for $x < 0$, so everywhere we have upward concavity. (It is rare for a point where $y"=0$ not to be an inflection point; this can occur only when the third derivative $y^{\prime\prime\prime}$ is also zero at the same point.)

\begin{centering} \begin{psfigure}{4.5in}{1.5in}{124.figNH.ps} \end{psfigure}\\ \end{centering}

\end{example}

\begin{example}

Sketch $y=x^5-5x^4+5x^3$.

First, we set $0=y'=5x^4-20x^3+15x^2$. To solve this, we factor out what we can, namely $5x^2$. This leaves a quadratic that can be factored either by inspection or by the quadratic formula. The result is $0=y'=5x^2(x-1)(x-3)$. Thus, the critical points are $(0,0)$, $(1,1)$, and $(3,-27)$. Using $y"=20x^3-60x^2+30x=10x(2x^2-6x+3)$, we see from the second derivative test that $(1,1)$ is a local maximum and $(3,-27)$ is a local minimum, but we get no information about $(0,0)$. Setting $0=y"$ and using the quadratic formula to find the roots of $2x^2-6x+3$, we find the following three points of inflection: $(0,0)$, $(0.634,0.569)$, $(2.366,-16.32)$.

\begin{rightindent}{3.0in}

In a complicated case like this, it is also worthwhile to see what the function is doing when $x$ is large positive or large negative. If $x$ is large, the $x^5$ term in our function dominates (is greater in absolute value than all the other terms). Thus, the function heads upward steeply into the first quadrant as $x\longrightarrow+\infty$, and it heads steeply down into the third quadrant as $x\longrightarrow-\infty$. Putting this information together, we obtain the graph shown above. The curve is concave downward in the third quadrant, and also between the two points of inflection $(0.634,0.569)$ and $(2.366,-16.32)$. For $0 < x < 0.634$ and for $x>2.366$ the curve is concave upward. \end{rightindent} \hfill \begin{psfigure}{2.45in}{2.5in}{124.figNI.ps} \end{psfigure}

\end{example}

\newpage

\begin{homework}\ \label{homework:13}

\medskip

\exercise Suppose that $n$ is an integer greater than 2. On the curve $y=f(x)=x^n$, what sort of point is the origin? Sketch the curve, and indicate the concavity.

\exercise In parts (a)-(j) below, find each max/min point (using the second derivative test to be sure what type of point it is), point of inflection, and asymptote (vertical, horizontal, or slanted), if there are any. Also indicate where the curve is concave up and concave down. Sketch the graph.

(a) $y=x^2-x$, (b) $y=2+3x-x^3$, (c) $y=x^3-9x^2+24x$, (d) $y=x^4-2x^2+3$, (e) $y=3x^4-4x^3$, (f) $y=(x^2-1)/x$, (g) $y=3x^2-(1/x^2)$, (h) $y=\cos(2x)-x$, (i) $y=\sin x+\cos x$, (j) $y=\tan(x/2)-x$.

\exercise Suppose $\begin{cases} f'(x) > 0& \text{for $|x| > 2$;}\\ f'(x) < 0& \text{for $|x| < 2$;}\\ f"(x) < 0& \text{for $x < 0$; and }\\ f"(x) > 0& \text{for $x > 0$}\end{cases}$.

From the given information, sketch a possible graph of $f(x)$. How could your answer vary and still be correct?

\exercise Suppose $f"(x) < 0$ for $x < 1$ and $f"(x)>0$ for $x>1$. Suppose also that $f(1)=1$. Sketch a possible graph of $f(x)$, assuming that

i) $f'(1) = 0$, ii) $f'(1) > 0$ and iii) $f'(1) < 0$.

\begin{rightindent}{3.4in} \exercise At the right is a sketch of $y=x^6-5x^4$.

(a) Find the exact coordinates $(x,y)$ of the local maxima and local minima.

(b) For what values of $x$ is the function concave upward? (c) Find the exact $x$-coordinates of all points of inflection. (Hint: Color the part of the curve that is concave up blue and color the part that is concave down red. Points of inflection occur only where the curve changes color!) \end{rightindent} \hfill \begin{psfigure}{2.0in}{2.0in}{124.figNJ.ps} \end{psfigure}

(d) Explain how you know from your calculations (not from the sketch you were given) which of your answers to part (a) are minima and which are maxima.

\end{homework}