91Ó°ÊÓ

It can be said that the domain of the given function \(f\left( x \right) = 1 + \frac{1}{x} - \frac{1}{{{x^2}}}\) is \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

(a).

For horizontal asymptote, find \(\mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \pm \infty } \left( {1 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)\\ = 1\end{array}\)

Thus, \(y = 1\) is a horizontal asymptote.

For vertical asymptote, find \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {1 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{{x^2} + x - 1}}{{{x^2}}}} \right)\\ = - \infty \end{array}\)

Hence, as \(x \to {0^ + }\), then \(\left( {{x^2} + x - 1} \right) \to - 1\) and as \({x^2} \to - 1\), then \(x \to {0^ - }\).

Thus, \(x = 0\) is the vertical asymptote.

ii) In an interval,\(f\)is increasing if \(f'\left( x \right) > 0\) on that interval.

(b).

Find the derivative of \(f\left( x \right) = 1 + \frac{1}{x} - \frac{1}{{{x^2}}}\) with respect to \(x\).

\(\begin{array}{c}f'\left( x \right) = \frac{d}{{dx}}\left( {1 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)\\ = - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}\\ = - \frac{1}{{{x^3}}}\left( {x - 2} \right)\end{array}\)

Furthermore, when \(f'\left( x \right) = 0\), \(x = 2\).

Draw a table for the interval of increasing and decreasing.

Assume a critical point \(c\) of a continuous function \(f\).

9. The function \(f\) has a local maximum at the point \(c\) if \(f'\) changes from positive to negative.
10. The function \(f\) has a local minimum at the point \(c\) if \(f'\) changes from negative to positive.
11. The function \(f\) does not have any local maximum or minimum at \(c\) if \(f'\) is positive/negative to the left and right of \(c\).

(c)

From the table in Step 2, it can be observed that \(f'\) changes from positive to negative at 2, so there will be a local maximum value at \(x = 2\).

At \(x = 2\),

\(\begin{array}{c}f\left( 2 \right) = 1 + \frac{1}{2} - \frac{1}{4}\\ = \frac{5}{4}\end{array}\)

So, the local maximum point is \(\left( {2,\frac{5}{4}} \right)\).

Concave upward on \(I\): if \(f''\left( x \right) > 0\) on interval \(I\)

Concave downward on \(I\): if \(f''\left( x \right) < 0\) on interval \(I\)

A point \(P\) is known as an inflection point if the function \(f\) is continuous and the curve changes from concave upward to concave downward or concave downward to concave upward at the point \(P\).

(d)

Find the double derivative of the function \(f\left( x \right) = 1 + \frac{1}{x} - \frac{1}{{{x^2}}}\).

\(\begin{array}{c}f''\left( x \right) = \frac{d}{{dx}}\left( { - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)\\ = \frac{2}{{{x^3}}} - \frac{6}{{{x^4}}}\\ = \frac{2}{{{x^4}}}\left( {x - 3} \right)\end{array}\)

Put \(f''\left( x \right) = 0\) and solve for \(x\).

\(\begin{array}{c}\frac{2}{{{x^4}}}\left( {x - 3} \right) = 0\\x = 3\end{array}\)

Draw a table for concavity for different intervals.

Find \(f\left( 3 \right)\) as from the table, you will get the inflection point.

\(\begin{array}{c}f\left( 3 \right) = 1 + \frac{1}{3} - \frac{1}{9}\\ = \frac{{11}}{9}\end{array}\)

So, the inflection point is \(\left( {3,\frac{{11}}{9}} \right)\).

(e)

Draw the graph of the given functions by using the obtained information.