Curve sketching I

Section 24 Curve sketching I

Objectives

Bring all our calculus tools to bear on the analysis of a real-valued function $f\text{.}$ In more detail: determine the domain, the $y$-intercept and any $x$-intercepts; compute the limit of the function at endpoints of the domain and determine whether there are horizontal asymptotes; determine whether there are vertical asymptotes; find and classify all critical points of $f\text{;}$ determine the intervals of monotonicity; determine intervals of constant concavity; identify any inflection points.
Express the results of our analysis of a function’s behavior in the form of a detailed sketch.

Procedure 24.1. Curve sketching.

When asked to give an accurate sketch of the graph of a function $f\text{,}$ we seek to produce a sketch that accurately reflects the following properties:

the domain $D$ of $f\text{;}$
$x$- and $y$-intercepts of $f\text{;}$
horizontal asymptotes, and more generally, the “endpoint” behavior of $f$ (i.e., limits of $f(x)$ as $x$ approaches endpoints of $D$);
vertical asymptotes;
critical points of $f\text{,}$ classified as points where $f$ attains a local maximum value, a local minimum value, or neither;
intervals of monotonicity of $f$ (i.e., intervals where $f$ is either increasing or decreasing);
intervals of constant concavity of $f\text{;}$
inflection points of $f\text{.}$

Example 24.2. Curve sketch: polynomial.

Provide a graph of $f(x)=3x^5-5x^3$ that includes all the details listed in Procedure 24.1.

Solution.

Domain and intercepts.

The domain of $f$ is $\R=(-\infty, \infty)\text{.}$ The $y$-intercept of the graph of $f$ is $(0,f(0))=(0,0)\text{.}$

To find the $x$-intercepts we solve:

\begin{align*} f(x) \amp =0\\ x^3(3x^2-5) \amp =0\\ x=0 \amp\text{ or } x^2=\frac{5}{3} \\ x=0 \amp\text{ or } x=\pm\sqrt{5/3} \text{.} \end{align*}

Thus the $x$-intercepts are

\begin{equation*} (0,0), (\sqrt{5/3},0), (-\sqrt{5/3},0)\text{.} \end{equation*}

Endpoint behavior and vertical asymptotes.

Since $f$ is continuous at all points of $\R\text{,}$ there are no vertical asymptotes.

For endpoint behavior, we compute

\begin{align*} \lim_{x\to \pm \infty}f(x) \amp = \lim\limits_{x\to \pm \infty}\frac{3x^5-5x^3}{1} \\ \amp = \lim\limits_{x\to \pm \infty}3x^5 \amp (\knowl{./knowl/xref/th_rational_function.html}{\text{Theorem 9.11}})\\ \amp = \pm \infty\text{.} \end{align*}

In particular, we see that there are no horizontal asymptotes of the graph of $f\text{.}$

Critical points and intervals of monotonicity.

We saw in Example 22.8 that the critical points of $f$ are $x=-1,0,1\text{,}$ that $f(-1)$ is a local maximum value, $f(1)$ is a local minimum value, and $f(0)$ is neither, that $f$ is increasing on the intervals $(-\infty, -1]$ and $[1,\infty)$ and decreasing on the interval $[-1,1]\text{.}$ All of this information is summarized by the sign diagram of $f'\text{.}$

$Sign diagram for derivative of f$

Concavity and inflection points.

We analyzed the concavity of $f$ in Example 23.9. The sign diagram we produced there gives a nice summary of the situation.

$Sign diagram for second derivative of f$

In particular, observe that we have inflection points at the inputs $x=-1/\sqrt{2}, 0, 1/\sqrt{2}\text{.}$

Important points.

Our analysis is more or less complete. We should first, however, make a table of values (to the best of our abilities) for all important points on our graph.

\begin{equation*} \begin{array}{r|l} x \amp f(x) \\ \hline 0\amp 0 \\ -1 \amp 2 \\ 1 \amp -2 \\ -\frac{1}{\sqrt{2}} \amp -(3(2)^{-5/2}+5(2)^{-3/2}) \\ \frac{1}{\sqrt{2}} \amp (3(2)^{-5/2}+5(2)^{-3/2}) \end{array}\text{.} \end{equation*}

Finally we put everything together into a single sketch.

Example 24.3. Curve sketching: rational function.

Provide a graph of $f(x)=\frac{x^2}{x^2+3}$ that includes all the details listed in Procedure 24.1.

Solution.

For the sake of space, our solution will be slightly more terse than in the previous example. We first observe that the domain of $f$ is $\R$ and that $(0,0)$ is both the $x$- and $y$-intercept.

Since $f$ is continuous everywhere on $\R\text{,}$ it has not vertical asymptotes. Using Theorem 9.11, we see easily that $\lim\limits_{x\to \pm \infty}f(x)=1\text{,}$ and thus that $y=1$ is a horizontal asymptote of $f\text{.}$

Next we compute

\begin{align*} f'(x) \amp = \frac{6x}{(x^2+3)^2}\\ f''(x) \amp = \frac{18(x-1)(x+1)}{(x^2+3)^3} \text{.} \end{align*}

The sign diagrams for $f'$ and $f''$ are given below.

$Sign diagram for derivative of f$

We see then that $f$ has a single critical point at $x=0\text{,}$ that $f(0)$ is a local minimum value of $f\text{,}$ and that $f$ is decreasing on $(-\infty, 0)$ and increasing on $(0,\infty)\text{.}$ Furthermore, $f$ has inflection points at $x=-1$ and $x=1\text{,}$ is concave down on $(-\infty, -1)$ and $(1,\infty)\text{,}$ and concave up on $(-1,1)\text{.}$

To finish our analysis we make a table of values of important points:

\begin{equation*} \begin{array}{r|l} x\amp f(x) \\ \hline 0 \amp 0 \\ -1 \amp \frac{1}{4} \\ 1 \amp \frac{1}{4} \end{array}\text{.} \end{equation*}

Putting everything together, we obtain a graph like the following.

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