# Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 50

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Use the guidelines of curve sketching to sketch the curve $\displaystyle xy = x^2 + x + 1$ then find the equation of slant asymptote.

A. Domain. We can rewrite $f(x)$ as $\displaystyle y = \frac{x^2 + x + 1}{x}$. Therefore, the domain of the function is $(- \infty, 0) \bigcup (0, \infty)$

B. Intercepts. Solving for $y$ intercept, when $x = 0$,

$\displaystyle y = \frac{0^2 + 0 + 1}{0}$, $y$ intercept does not exist

Solving for $x$ intercept, when $y = 0$,

\begin{aligned} 0 =& \frac{x^2 + x + 1}{x} \\ \\ 0 =& x^2 + x + 1 \end{aligned}

$x$ intercept does not exist

C. Symmetry. The function is not symmetric to either $y$ axis and origin by using symmetry test.

D. Asymptote. For vertical asymptote, we have $x = 0$

For horizontal asymptote, since $\displaystyle \lim_{x \to \pm \infty} f(x) = \pm \infty$, the function has no horizontal asymptote.

For slant asymptote, by using long division,

We can rewrite $f(x)$ as $\displaystyle y = x + 1 + \frac{1}{x}$, so...

$\displaystyle \lim_{x \to \pm \infty} f(x) - (x + 1) = \frac{1}{x} = 0$

Therefore, the equation of slant asymptote is $y = x + 1$.

E. Intervals of increase or decrease,

If $\displaystyle f(x) = \frac{x^2 + x + 1}{x}$, then by using Quotient Rule,

\begin{aligned} f'(x) =& \frac{x (2x + 1) - (x^2 + x + 1)(1)}{x^2} = \frac{2x^2 - \cancel{x} - x^2 - \cancel{x} - 1}{x^2} = \frac{x^2 - 1}{x^2} \\ \\ f'(x) =& 1 - \frac{1}{x^2} \end{aligned}

When $f'(x) = 0$

\begin{aligned} 0 =& 1 - \frac{1}{x^2} \\ \\ \frac{1}{x^2} =& 1 \end{aligned}

The critical numbers are

$x = 1$ and $x = -1$

Hence, the intervals of increase and decrease are..

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f \\ x < -1 & + & \text{increasing on } (- \infty, -1) \\ -1 < x < 1 & - & \text{decreasing on$(-1, 1)$except at$x = 0$} \\ 2 < x < 6 & - & \text{decreasing on (2, 6)} \\ x > 1 & + & \text{increasing on } (6, \infty) \\ \hline \end{array}$

F. Local Maximum and Minimum Values

Since $f'(x)$ changes from positive to negative at $x = -1, f(-1) = -1$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive at $x = 1, f(1) = 3$ is a local minimum.

G. Concavity and inflection point

If $f'(x) = \displaystyle 1 - \frac{1}{(x^2)}$, then

\begin{aligned} f''(x) =& \frac{2}{x^3} \end{aligned}

when $f''(x) = 0,$

$\displaystyle 0 = \frac{2}{x^3}$

$f''(x)= 0$ does not exist, therefore the function has no inflection point.

Hence, the concavity is..

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity} \\ x < 0 & - & \text{Downward} \\ x > 0 & + & \text{Upward}\\ \hline \end{array}$

H. Sketch the graph.