Question

17–22 Use a computer algebra system to graph \(f\) and to find \(f'\) and \(f''\). Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of \(f\).

17. \(f\left( x \right) = \frac{{{x^3} + 5{x^2} + 1}}{{{x^4} + {x^3} - {x^2} + 2}}\)

Short answer

\(f'\left( x \right) = \frac{{ - x\left( {{x^5} + 10{x^4} + 6{x^3} + 4{x^2} - 3x - 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^2}}}\)

\(f''\left( x \right) = \frac{{2\left( {{x^9} + 15{x^8} + 18{x^7} + 21{x^6} - 9{x^5} - 135{x^4} - 76{x^3} + 21{x^2} + 6x + 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^3}}}\)

Decreasing Intervals: \(\left( { - \infty , - 9.41} \right)\), \(\left( { - 1.29,0} \right)\) and \(\left( {1.05,\infty } \right)\)

Increasing intervals: \(\left( { - 9.41, - 1.29} \right)\) and \(\left( {0,1.05} \right)\)

Local maximum values: \(\left( { - 1.29,7.49} \right)\)and\(\left( {1.05,2.35} \right)\)

Local minimum values: \(\left( { - 0.941, - 0.056} \right)\)and\(\left( {0,0.5} \right)\)

Concave up intervals: \(\left( { - 13.81, - 1.55} \right)\), \(\left( { - 1.03,0.60} \right)\) and \(\left( {1.48,\infty } \right)\)

Concave down intervals: \(\left( { - \infty , - 13.81} \right)\), \(\left( { - 1.55, - 1.03} \right)\) and \(\left( {0.60,1.48} \right)\)

Inflection points: \(\left( { - 13.81, - 0.05} \right)\), \(\left( {1.55,5.64} \right)\), \(\left( { - 1.03,5.39} \right)\), \(\left( {0.60,1.52} \right)\) and \(\left( {1.48,1.93} \right)\)

Step-by-step solution

Determine the first derivative of the function

The given function is \(f\left( x \right) = \frac{{{x^3} + 5{x^2} + 1}}{{{x^4} + {x^3} - {x^2} + 2}}\).

Use the following steps to find the derivative of \(f\left( x \right)\) in Wolframalpha.com.

1. Enter the function \(f\left( x \right) = \frac{{{x^3} + 5{x^2} + 1}}{{{x^4} + {x^3} - {x^2} + 2}}\) in the tab in the form of \(\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{{x^3} + 5{x^2} + 1}}{{{x^4} + {x^3} - {x^2} + 2}}} \right)\).
2. Press the \( = \) button on the right side of tab.

So, the derivative of \(f\left( x \right)\) is:

Use an online calculator to get the first derivative of the function that is \(f'\left( x \right)\).

\(f'\left( x \right) = \frac{{ - x\left( {{x^5} + 10{x^4} + 6{x^3} + 4{x^2} - 3x - 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^2}}}\)

Determine the second derivative of the function

Differentiate the function again using the Wolfarmalpha.com calculator.

1. Enter the function \(f'\left( x \right) = \frac{{ - x\left( {{x^5} + 10{x^4} + 6{x^3} + 4{x^2} - 3x - 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^2}}}\) in the tab in the form of \(\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{ - x\left( {{x^5} + 10{x^4} + 6{x^3} + 4{x^2} - 3x - 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^2}}}} \right)\).
2. Press the \( = \) button on the right side of the tab.

So, the derivative of \(f'\left( x \right)\) is:

.

\(f''\left( x \right) = \frac{{2\left( {{x^9} + 15{x^8} + 18{x^7} + 21{x^6} - 9{x^5} - 135{x^4} - 76{x^3} + 21{x^2} + 6x + 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^3}}}\)

Graphs of \(f\left( x \right)\), \(f'\left( x \right)\) and \(f''\left( x \right)\)

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer, visually draw the graph of the functions\(f\left( x \right) = \frac{{{x^3} + 5{x^2} + 1}}{{{x^4} + {x^3} - {x^2} + 2}}\),\(f'\left( x \right) = \frac{{ - x\left( {{x^5} + 10{x^4} + 6{x^3} + 4{x^2} - 3x - 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^2}}}\)and\(f''\left( x \right) = \frac{{2\left( {{x^9} + 15{x^8} + 18{x^7} + 21{x^6} - 9{x^5} - 135{x^4} - 76{x^3} + 21{x^2} + 6x + 22} \right)}}{{{{\left( {{x^4} + {x^3} - {x^2} + 2} \right)}^3}}}\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\left( {{X^3} + 5{X^2} + 1} \right)/\left( {{X^4} + {X^3} - {X^2} + 2} \right)\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function\(f'\left( x \right)\)is shown below:

For\(f'\left( x \right)\):

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\( - X\left( {{X^5} + 10{X^4} + 6{X^3} + 4{X^2} - 3X - 22} \right)/{\left( {{X^4} + {X^3} - {X^2} + 2} \right)^2}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f'\left( x \right)\) is shown below:

For\(f''\left( x \right)\):

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\begin{aligned}{l}2\left( {{X^9} + 15{X^8} + 18{X^7} + 21{X^6} - 9{X^5} - 135{X^4} - 76{X^3} + 21{X^2} + 6X + 22} \right)/\\{\left( {{X^4} + {X^3} - {X^2} + 2} \right)^3}\end{aligned}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f''\left( x \right)\) is shown below:

Determine increase decrease intervals

Consider the graph of\(f'\left( x \right)\).

From the graph, it can be concluded that there are four zeros and\(f\)is decreasing on \(\left( { - \infty , - 9.41} \right)\), \(\left( { - 1.29,0} \right)\) and \(\left( {1.05,\infty } \right)\).

The function is increasing on \(\left( { - 9.41, - 1.29} \right)\), and \(\left( {0,1.05} \right)\).

Determine local minimum and maximum values

From the graph of \(f\), it can be concluded that there is an \(x\)-intercept near \( - 5\), and the graph of the function will have a local minimum and inflection point of the left of \( - 5\).

And from the graph of \(f'\left( x \right)\), itcan be concluded that there is a local minimum value at\(x = - 9.41,0\)and local maximum value at\(x = - 1.29,1.05\).

By using an online calculator,\(f\left( { - 0.941} \right) \approx - 0.056\),\(f\left( 0 \right) = 0.5\),\(f\left( { - 1.29} \right) \approx 7.49\)and\(f\left( {1.05} \right) \approx 2.35\).

So, local minimum points are,\(\left( { - 0.941, - 0.056} \right)\)and\(\left( {0,0.5} \right)\). Local maximum points are\(\left( { - 1.29,7.49} \right)\)and\(\left( {1.05,2.35} \right)\).

Determine intervals of concavity and inflection points

Consider the graph of\(f''\left( x \right)\).

From the graph, it can be concluded that there are five zeros, so there will be five inflection points. Where,\(f\)is concave up on \(\left( { - 13.81, - 1.55} \right)\), \(\left( { - 1.03,0.60} \right)\) and \(\left( {1.48,\infty } \right)\).

The function is concave down on \(\left( { - \infty , - 13.81} \right)\), \(\left( { - 1.55, - 1.03} \right)\) and \(\left( {0.60,1.48} \right)\).

And the inflection points from the graph are, \(\left( { - 13.81, - 0.05} \right)\), \(\left( {1.55,5.64} \right)\), \(\left( { - 1.03,5.39} \right)\), \(\left( {0.60,1.52} \right)\) and \(\left( {1.48,1.93} \right)\).