Question

2. For the following rational functions: (1) First sketch a rough graph of the function using superposition of simple functions. (2) Find the coordinates of the local max/min using $f'$ and add them to the plot. (3) Find the coordinates of all inflection points using $f''$ and add them to the plot. (a) $f(x) = x + \frac{1}{x}$, $x > 0$ (b) $f(x) = x + \frac{1}{x^2}$, $x > 0$

          2. For the following rational functions:
(1) First sketch a rough graph of the function using superposition of simple functions.
(2) Find the coordinates of the local max/min using $f'$ and add them to the plot.
(3) Find the coordinates of all inflection points using $f''$ and add them to the plot.
(a) $f(x) = x + \frac{1}{x}$, $x > 0$
(b) $f(x) = x + \frac{1}{x^2}$, $x > 0$

2. For the following rational functions:
(1) First sketch a rough graph of the function using superposition of simple functions.
(2) Find the coordinates of the local max/min using f' and add them to the plot.
(3) Find the coordinates of all inflection points using f” and add them to the plot.
(a) f(x) = x + (1)/(x), x > 0
(b) f(x) = x + (1)/(x^2), x > 0

Added by Irene R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zack A

10/16/2023

Step 1

$f'(x) = 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1$ (since $x > 0$). $f''(x) = \frac{2}{x^3}$. $f''(1) = 2 > 0$, so $x = 1$ is a local minimum. $f(1) = 1 + 1 = 2$. The local minimum is at $(1, 2)$. There are no inflection points since $f''(x) > 0$ for Show more…

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2. For the following rational functions: (1) First sketch a rough graph of the function using superposition of simple functions. (2) Find the coordinates of the local max/min using $f'$ and add them to the plot. (3) Find the coordinates of all inflection points using $f''$ and add them to the plot. (a) $f(x) = x + \frac{1}{x}$, $x > 0$ (b) $f(x) = x + \frac{1}{x^2}$, $x > 0$