Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 16
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Use the guidelines of curve sketching to sketch the curve. $\displaystyle y =1 + \frac{1}{x} + \frac{1}{x^2}$
The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a rational function that is defined everywhere except for the value of $x$ that would make its denominator equal to 0. In our case, $x=0$. Therefore, the domain is $(-\infty,0)\bigcup(0,\infty)$
B. Intercepts.
We don't have $y$-intercept since zero is not included in the domain of $f$.
Solving for $x$-intercept, when $y = 0$
$\displaystyle 0 = \frac{x^2 + x + 1}{x^2}$
$0 = x^2 + x + 1$
We have no real solution for this. Hence, there is no $x$-intercept.
C. Symmetry.
The function is not symmetric in either $y$-axis or origin.
D. Asymptotes.
For vertical asymptotes, we set the denominator equal to 0, that is $x^2 = 0$ therefore, $x = 0$ is our vertical asymptote.
For horizontal asymptotes, we divide the coefficient of the highest degree of the numerator and denominator to obtain $\displaystyle y= \frac{1}{1} = 1$
E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule.
$\displaystyle f'(x) = \frac{x^2(2x+1)-(x^2+x+1)(2x)}{(x^2)^2} = \frac{-2-x}{x^3}$
When $f'(x) = 0$,
$\displaystyle 0 = \frac{-2-x}{x^3}$
We have, $x = -2$ as our critical number
If we divide the interval, we can determine the intervals of increase or decrease,
$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < - 2 & - & \text{decreasing on } (-\infty, -2)\\ \hline\\ x > -2 & + & \text{increasing on } (-2, \infty)\\ \hline \end{array} $
F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from negative to positive at $x = -2$ , $\displaystyle f(-2) = \frac{3}{4}$ is a local minimum.
G. Concavity and Points of Inflection.
$ \begin{equation} \begin{aligned} \text{if } f'(x) &= \frac{-2-x}{x^3}, \text{ then}\\ \\ f''(x) &= \frac{x^3 ( - 1) - ( - 2 - x ) (3x^2)}{(x^3)^2} \end{aligned} \end{equation} $
Which can be simplified as, $\displaystyle f''(x) = \frac{2(x+3)}{x^4}$
When $f''(x) = 0$
$0 = 2 ( x+3)\\ x + 3 = 0$
The inflection is at $x = -3$
If we divide the interval, we can determine the concavity as,
$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity}\\ \hline\\ x < - 3 & - & \text{Downward}\\ \hline\\ x > -3 & + & \text{Upward}\\ \hline \end{array} $
H. Sketch the Graph.
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