4.6 Limits at infinity and asymptotes  (Page 7/14)

Step 1: Since $f$ is a polynomial, the domain is the set of all real numbers.

Step 2: When $x=0,f\left(x\right)=2.$ Therefore, the $y$ -intercept is $\left(0,2\right).$ To find the $x$ -intercepts, we need to solve the equation ${\left(x-1\right)}^2\left(x+2\right)=0,$ gives us the $x$ -intercepts $\left(1,0\right)$ and $\left(\mathrm{-2},0\right)$

Step 3: We need to evaluate the end behavior of $f.$ As $x\to \infty ,$ ${\left(x-1\right)}^2\to \infty$ and $\left(x+2\right)\to \infty .$ Therefore, $\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty .$ As $x\to \text{−}\infty ,$ ${\left(x-1\right)}^2\to \infty$ and $\left(x+2\right)\to \text{−}\infty .$ Therefore, $\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty .$ To get even more information about the end behavior of $f,$ we can multiply the factors of $f.$ When doing so, we see that

Since the leading term of $f$ is $x^3,$ we conclude that $f$ behaves like $y=x^3$ as $x\to \text{±}\infty .$

Step 4: Since $f$ is a polynomial function, it does not have any vertical asymptotes.

Step 5: The first derivative of $f$ is

Therefore, $f$ has two critical points: $x=1,\mathrm{-1}.$ Divide the interval $\left(\text{−}\infty ,\infty \right)$ into the three smaller intervals: $\left(\text{−}\infty ,\mathrm{-1}\right),$ $\left(\mathrm{-1},1\right),$ and $\left(1,\infty \right).$ Then, choose test points $x=\mathrm{-2},$ $x=0,$ and $x=2$ from these intervals and evaluate the sign of $f^{\prime }\left(x\right)$ at each of these test points, as shown in the following table.

From the table, we see that $f$ has a local maximum at $x=\mathrm{-1}$ and a local minimum at $x=1.$ Evaluating $f\left(x\right)$ at those two points, we find that the local maximum value is $f\left(\mathrm{-1}\right)=4$ and the local minimum value is $f\left(1\right)=0.$

Step 6: The second derivative of $f$ is

The second derivative is zero at $x=0.$ Therefore, to determine the concavity of $f,$ divide the interval $\left(\text{−}\infty ,\infty \right)$ into the smaller intervals $\left(\text{−}\infty ,0\right)$ and $\left(0,\infty \right),$ and choose test points $x=\mathrm{-1}$ and $x=1$ to determine the concavity of $f$ on each of these smaller intervals as shown in the following table.

We note that the information in the preceding table confirms the fact, found in step $5,$ that $f$ has a local maximum at $x=\mathrm{-1}$ and a local minimum at $x=1.$ In addition, the information found in step $5$ —namely, $f$ has a local maximum at $x=\mathrm{-1}$ and a local minimum at $x=1,$ and $f^{\prime }\left(x\right)=0$ at those points—combined with the fact that $f\text{″}$ changes sign only at $x=0$ confirms the results found in step $6$ on the concavity of $f.$

Combining this information, we arrive at the graph of $f\left(x\right)={\left(x-1\right)}^2\left(x+2\right)$ shown in the following graph.

Step 1: The function $f$ is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers $x$ except $x=\text{±}1.$

Step 2: Find the intercepts. If $x=0,$ then $f\left(x\right)=0,$ so $0$ is an intercept. If $y=0,$ then $\frac{x^2}{\left(1-x^2\right)}=0,$ which implies $x=0.$ Therefore, $\left(0,0\right)$ is the only intercept.

Step 3: Evaluate the limits at infinity. Since $f$ is a rational function, divide the numerator and denominator by the highest power in the denominator: $x^2.$ We obtain

Therefore, $f$ has a horizontal asymptote of $y=\mathrm{-1}$ as $x\to \infty$ and $x\to \text{−}\infty .$

Step 4: To determine whether $f$ has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when $x=\text{±}1.$ To determine whether the lines $x=1$ or $x=\mathrm{-1}$ are vertical asymptotes of $f,$ evaluate $\underset{x\to 1}{\text{lim}}f\left(x\right)$ and $\underset{x\to \text{−}1}{\text{lim}}f\left(x\right).$ By looking at each one-sided limit as $x\to 1,$ we see that

In addition, by looking at each one-sided limit as $x\to \text{−}1,$ we find that

Step 5: Calculate the first derivative:

Critical points occur at points $x$ where $f^{\prime }\left(x\right)=0$ or $f^{\prime }\left(x\right)$ is undefined. We see that $f^{\prime }\left(x\right)=0$ when $x=0.$ The derivative $f^{\prime }$ is not undefined at any point in the domain of $f.$ However, $x=\text{±}1$ are not in the domain of $f.$ Therefore, to determine where $f$ is increasing and where $f$ is decreasing, divide the interval $\left(\text{−}\infty ,\infty \right)$ into four smaller intervals: $\left(\text{−}\infty ,\mathrm{-1}\right),$ $\left(\mathrm{-1},0\right),$ $\left(0,1\right),$ and $\left(1,\infty \right),$ and choose a test point in each interval to determine the sign of $f^{\prime }\left(x\right)$ in each of these intervals. The values $x=\mathrm{-2},$ $x=-\frac{1}{2},$ $x=\frac{1}{2},$ and $x=2$ are good choices for test points as shown in the following table.

From this analysis, we conclude that $f$ has a local minimum at $x=0$ but no local maximum.

Step 6: Calculate the second derivative:

To determine the intervals where $f$ is concave up and where $f$ is concave down, we first need to find all points $x$ where $f\text{″}\left(x\right)=0$ or $f\text{″}\left(x\right)$ is undefined. Since the numerator $6x^2+2\ne 0$ for any $x,$ $f\text{″}\left(x\right)$ is never zero. Furthermore, $f\text{″}$ is not undefined for any $x$ in the domain of $f.$ However, as discussed earlier, $x=\text{±}1$ are not in the domain of $f.$ Therefore, to determine the concavity of $f,$ we divide the interval $\left(\text{−}\infty ,\infty \right)$ into the three smaller intervals $\left(\text{−}\infty ,\mathrm{-1}\right),$ $\left(\mathrm{-1},\mathrm{-1}\right),$ and $\left(1,\infty \right),$ and choose a test point in each of these intervals to evaluate the sign of $f\text{″}\left(x\right).$ in each of these intervals. The values $x=\mathrm{-2},$ $x=0,$ and $x=2$ are possible test points as shown in the following table.

Combining all this information, we arrive at the graph of $f$ shown below. Note that, although $f$ changes concavity at $x=\mathrm{-1}$ and $x=1,$ there are no inflection points at either of these places because $f$ is not continuous at $x=\mathrm{-1}$ or $x=1.$