4.6 Limits at infinity and asymptotes  (Page 8/14)

Step 1: The domain of $f$ is the set of all real numbers $x$ except $x=1.$

Step 2: Find the intercepts. We can see that when $x=0,$ $f\left(x\right)=0,$ so $\left(0,0\right)$ is the only intercept.

Step 3: Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, $f$ must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write

Since $1\text{/}\left(x-1\right)\to 0$ as $x\to \text{±}\infty ,$ $f\left(x\right)$ approaches the line $y=x+1$ as $x\to \text{±}\infty .$ The line $y=x+1$ is an oblique asymptote for $f.$

Step 4: To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at $x=1.$ Looking at both one-sided limits as $x\to 1,$ we find

Therefore, $x=1$ is a vertical asymptote, and we have determined the behavior of $f$ as $x$ approaches $1$ from the right and the left.

Step 5: Calculate the first derivative:

We have $f^{\prime }\left(x\right)=0$ when $x^2-2x=x\left(x-2\right)=0.$ Therefore, $x=0$ and $x=2$ are critical points. Since $f$ is undefined at $x=1,$ we need to divide the interval $\left(\text{−}\infty ,\infty \right)$ into the smaller intervals $\left(\text{−}\infty ,0\right),$ $\left(0,1\right),$ $\left(1,2\right),$ and $\left(2,\infty \right),$ and choose a test point from each interval to evaluate the sign of $f^{\prime }\left(x\right)$ in each of these smaller intervals. For example, let $x=\mathrm{-1},$ $x=\frac{1}{2},$ $x=\frac{3}{2},$ and $x=3$ be the test points as shown in the following table.

From this table, we see that $f$ has a local maximum at $x=0$ and a local minimum at $x=2.$ The value of $f$ at the local maximum is $f\left(0\right)=0$ and the value of $f$ at the local minimum is $f\left(2\right)=4.$ Therefore, $\left(0,0\right)$ and $\left(2,4\right)$ are important points on the graph.

Step 6: Calculate the second derivative:

We see that $f\text{″}\left(x\right)$ is never zero or undefined for $x$ in the domain of $f.$ Since $f$ is undefined at $x=1,$ to check concavity we just divide the interval $\left(\text{−}\infty ,\infty \right)$ into the two smaller intervals $\left(\text{−}\infty ,1\right)$ and $\left(1,\infty \right),$ and choose a test point from each interval to evaluate the sign of $f\text{″}\left(x\right)$ in each of these intervals. The values $x=0$ and $x=2$ are possible test points as shown in the following table.

From the information gathered, we arrive at the following graph for $f.$

Step 1: Since the cube-root function is defined for all real numbers $x$ and ${\left(x-1\right)}^{2\text{/}3}={\left(\sqrt[3]{x-1}\right)}^2,$ the domain of $f$ is all real numbers.

Step 2: To find the $y$ -intercept, evaluate $f\left(0\right).$ Since $f\left(0\right)=1,$ the $y$ -intercept is $\left(0,1\right).$ To find the $x$ -intercept, solve ${\left(x-1\right)}^{2\text{/}3}=0.$ The solution of this equation is $x=1,$ so the $x$ -intercept is $\left(1,0\right).$

Step 3: Since $\underset{x\to \text{±}\infty }{\text{lim}}{\left(x-1\right)}^{2\text{/}3}=\infty ,$ the function continues to grow without bound as $x\to \infty$ and $x\to \text{−}\infty .$

Step 4: The function has no vertical asymptotes.

Step 5: To determine where $f$ is increasing or decreasing, calculate $f^{\prime }.$ We find

This function is not zero anywhere, but it is undefined when $x=1.$ Therefore, the only critical point is $x=1.$ Divide the interval $\left(\text{−}\infty ,\infty \right)$ into the smaller intervals $\left(\text{−}\infty ,1\right)$ and $\left(1,\infty \right),$ and choose test points in each of these intervals to determine the sign of $f^{\prime }\left(x\right)$ in each of these smaller intervals. Let $x=0$ and $x=2$ be the test points as shown in the following table.

We conclude that $f$ has a local minimum at $x=1.$ Evaluating $f$ at $x=1,$ we find that the value of $f$ at the local minimum is zero. Note that $f^{\prime }\left(1\right)$ is undefined, so to determine the behavior of the function at this critical point, we need to examine $\underset{x\to 1}{\text{lim}}{f}^{\prime }\left(x\right).$ Looking at the one-sided limits, we have

Therefore, $f$ has a cusp at $x=1.$

Step 6: To determine concavity, we calculate the second derivative of $f\text{:}$

We find that $f\text{″}\left(x\right)$ is defined for all $x,$ but is undefined when $x=1.$ Therefore, divide the interval $\left(\text{−}\infty ,\infty \right)$ into the smaller intervals $\left(\text{−}\infty ,1\right)$ and $\left(1,\infty \right),$ and choose test points to evaluate the sign of $f\text{″}\left(x\right)$ in each of these intervals. As we did earlier, let $x=0$ and $x=2$ be test points as shown in the following table.

From this table, we conclude that $f$ is concave down everywhere. Combining all of this information, we arrive at the following graph for $f.$