Question

9-10 Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly.

9. \(f\left( x \right) = {\bf{1}} + \frac{{\bf{1}}}{x} + \frac{{\bf{8}}}{{{x^{\bf{2}}}}} + \frac{{\bf{1}}}{{{x^{\bf{3}}}}}\)

Short answer

The graph is shown below:

It appears that f increases on \(\left( { - 15.8, - 0.2} \right)\) and decrease on \(\left( { - \infty , - 15.8} \right)\), \(\left( { - 0.2,0} \right)\), and \(\left( {0,\infty } \right)\). The function f has a local minimum value at \(x = - 15.8\) , and the minimum value is 0.97. The local maximum value at \(x = - 0.2\) and the maximum value is 72.

The curve of the function f is concave downward on \(\left( { - \infty , - 24} \right)\) and \(\left( { - 0.25,0} \right)\) and concave upward on \(\left( { - 24, - 0.25} \right)\) and \(\left( {0,\infty } \right)\).

The inflection points are \(\left( { - 24,0.97} \right)\) and \(\left( { - 0.25,60} \right)\).

The curve of the function f is concave upward on \(\left( { - 12 - \sqrt {138} , - 12 + \sqrt {138} } \right)\) and \(\left( {0,\infty } \right)\) and \(f''\) is concave downward on \(\left( { - \infty , - 12 - \sqrt {138} } \right)\) and \(\left( { - 12 + \sqrt {138} ,0} \right)\).

Step-by-step solution

Differentiate the function \(f\left( x \right)\)

Differentiate the function \(f\left( x \right) = 1 + \frac{1}{x} + \frac{8}{{{x^2}}} + \frac{1}{{{x^3}}}\).

\(\begin{array}{c}f'\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {1 + \frac{1}{x} + \frac{8}{{{x^2}}} + \frac{1}{{{x^3}}}} \right)\\ = 0 - \frac{1}{{{x^2}}} - \frac{{16}}{{{x^3}}} - \frac{3}{{{x^4}}}\\ = - \frac{1}{{{x^4}}}\left( {{x^2} + 16x + 3} \right)\end{array}\)

Differentiate the function \(f'\left( x \right) = - \frac{1}{{{x^4}}}\left( {{x^2} + 16x + 3} \right)\).

\(\begin{array}{c}f''\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( { - \frac{1}{{{x^4}}}\left( {{x^2} + 16x + 3} \right)} \right)\\ = - \frac{{{x^4}\left( {2x + 16} \right) - \left( {{x^2} + 16x + 3} \right)\left( {4{x^3}} \right)}}{{{{\left( {{x^4}} \right)}^2}}}\\ = - \frac{1}{{{x^5}}}\left( {2{x^2} + 16x - 4{x^2} - 64 - 12} \right)\\ = \frac{2}{{{x^{\bf{5}}}}}\left( {{x^2} + 24x + 6} \right)\end{array}\)

Sketch the graph of f, \(f'\), and \(f''\)

Use the following steps to plot the graph of given functions:

1. In the graphing calculator, select “STAT PLOT” and enter the equations \(1 + \frac{1}{x} + \frac{8}{{{x^2}}} + \frac{1}{{{x^3}}}\) in the \({Y_1}\) tab.
2. Enter the graph button in the graphing calculator.

The figure below represents the graph of \(f\left( x \right)\).

\

Write an interpretation of graphs of f

It appears that f increases on \(\left( { - 15.8, - 0.2} \right)\) and decrease on \(\left( { - \infty , - 15.8} \right)\), \(\left( { - 0.2,0} \right)\), and \(\left( {0,\infty } \right)\). The function f has a local minimum value at \(x = - 15.8\) , and the minimum value is 0.97. The local maximum value at \(x = - 0.2\) and the maximum value is 72.

The curve of the function f is concave downward on \(\left( { - \infty , - 24} \right)\) and \(\left( { - 0.25,0} \right)\) and is concave upward on \(\left( { - 24, - 0.25} \right)\) and \(\left( {0,\infty } \right)\).

The inflection points are \(\left( { - 24,0.97} \right)\) and \(\left( { - 0.25,60} \right)\).

Find the exact value intervals

The root of the equation \(f'\left( x \right) = 0\) can be calculated as follows:

\(\begin{array}{c} - \frac{1}{{{x^4}}}\left( {{x^2} + 16x + 3} \right) = 0\\x = \frac{{ - 16 \pm \sqrt {256 - 12} }}{2}\\ = - 8 \pm \sqrt {61} \\ \approx - 0.19,\, - 15.81\end{array}\)

Here, \(f'\left( x \right) > 0\) on \(\left( { - 8 - \sqrt {61} , - 8 + \sqrt {61} } \right)\), \(f'\left( x \right) < 0\) on \(\left( { - \infty , - 8 - \sqrt {61} } \right)\), \(\left( { - 8 + \sqrt {61} ,0} \right)\) and \(\left( {0,\infty } \right)\).

Find the root of the equation \(f''\left( x \right) = 0\) can be calculated as follows:

\(\begin{array}{c}\frac{2}{{{x^{\bf{5}}}}}\left( {{x^2} + 24x + 6} \right) = 0\\x = \frac{{ - 24 \pm \sqrt {{{\left( {24} \right)}^2} - 4\left( 6 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}\\ = - 12 \pm \sqrt {138} \\ \approx - 0.25,\, - 23.75\end{array}\)

So, the curve of the function f is concave upward on \(\left( { - 12 - \sqrt {138} , - 12 + \sqrt {138} } \right)\) and \(\left( {0,\infty } \right)\) and \(f''\) is concave downward on \(\left( { - \infty , - 12 - \sqrt {138} } \right)\) and \(\left( { - 12 + \sqrt {138} ,0} \right)\).