Question

13-14 Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs using a calculator or computer that display the major features of the curve. Use these graphs to estimate the maximum and minimum values.

13. \(f\left( x \right) = \frac{{\left( {x + {\bf{4}}} \right)\,{{\left( {x - {\bf{3}}} \right)}^{\bf{2}}}}}{{{x^{\bf{4}}}\left( {x - {\bf{1}}} \right)}}\)

Short answer

The graph is shown below:

From the above graphs, it can be observed that there are three maximum values and one minimum value. The maximum values are \(f\left( { - 5.6} \right) = 0.0182\),\(f\left( {0.82} \right) = - 281.5\) and \(f\left( {5.2} \right) = 0.0145\), and the minimum value is \(f\left( 3 \right) = 0\).

Step-by-step solution

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

For the function \(f\left( x \right)\), \(x \to 0\), \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = - \infty \). So, \(x = 0\) is a vertical asymptote.

Similarly, as \(x \to {1^ - }\), \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = - \infty \).

And, as \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \infty \).

So, \(x = 1\) is also a vertical asymptote.

The function \(f\left( x \right) = \frac{{\left( {x + 4} \right){{\left( {x - 3} \right)}^2}}}{{{x^4}\left( {x - 1} \right)}}\) can be simplified as follows:

\(\begin{array}{c}f\left( x \right) = \frac{{\left( {x + 4} \right){{\left( {x - 3} \right)}^2}}}{{{x^4}\left( {x - 1} \right)}}\\ = \frac{{\frac{{x + 4}}{x} \cdot \frac{{{{\left( {x - 3} \right)}^2}}}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^3}}} \cdot \left( {x - 1} \right)}}\\ = \frac{{\left( {1 + \frac{4}{x}} \right){{\left( {1 - \frac{3}{x}} \right)}^2}}}{{x\left( {x - 1} \right)}}\\\mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) = 0\end{array}\)

So, \(y = 0\) is a horizontal asymptote to the curve.

The x-intercepts of the curve can be calculated by the roots of \(f\left( x \right) = 0\).

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

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

Find the maximum and minimum value of the function

The function is not defined at \(x = 0\), and there is no y-intercept. The only tangent of the graph is the x-axis, and it doesn’t cross it at \(x = 3\)since fis positive as \(x \to {3^ - }\) and \(x \to {3^ + }\).

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

1. In the graphing calculator, select “STAT PLOT” and enter the equations \(\frac{{\left( {x + 4} \right){{\left( {x - 3} \right)}^2}}}{{{x^4}\left( {x - 1} \right)}}\).
2. Set the window size \( - 8 \le X \le - 3.5\), and \( - 0.04 \le Y \le 0.02\).
3. Enter the graph button in the graphing calculator.

Set the window size \( - 1 \le X \le 2\), and \( - 1500 \le Y \le 500\).

Set the window size \(2.5 \le X \le 8\), and \(0 \le Y \le 0.03\).

From the above graphs, it can be observed that there are three maximum values and one minimum value. The maximum values are \(f\left( { - 5.6} \right) = 0.0182\),\(f\left( {0.82} \right) = - 281.5\) and \(f\left( {5.2} \right) = 0.0145\), and the minimum value is \(f\left( 3 \right) = 0\).