Question

11-12

1. Graph the function.
2. Use L’Hospital’s Rule to explain the behavior \(x \to {\bf{0}}\).
3. Estimate the minimum value and intervals of concavity. Then use calculus to find the exact values.

11. \(f\left( x \right) = {x^{\bf{2}}}{\bf{ln}}\,x\)

Short answer

a. The graph of the function is shown below:

b.The value using L’Hospital’s rule is 0.

c. The function f is increasing on \(\left( {\frac{1}{{\sqrt e }},\infty } \right)\) and decreasing on \(\left( {0,\frac{1}{{\sqrt e }}} \right)\). By the first derivative test, the local minimum value of the function is \(f\left( {\frac{1}{{\sqrt e }}} \right) = \frac{1}{{2e}}\). The point is approximate \(\left( { - 0.6065,0.1839} \right)\). The curve of f is concave upward on \(\left( {{e^{ - \frac{3}{2}}},\infty } \right)\) and concave downward on \(\left( {0,{e^{ - \frac{3}{2}}}} \right)\). The inflection point is \(\left( {{e^{ - \frac{3}{2}}},\frac{{ - 3}}{{2{e^3}}}} \right)\).

Step-by-step solution

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

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

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

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

The domain of \(f\left( x \right)\) is \(\left( {0,\infty } \right)\).

Apply L’Hospital’s Rule

The L’Hospital’s rule for the function \(f\left( x \right) = {x^2}\ln x\) can be applied as:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ + }} {x^2}\ln x = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\left( {\frac{1}{{{x^2}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left( {\frac{1}{x}} \right)}}{{\left( { - \frac{2}{{{x^3}}}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{L'Hospital's Rule}}} \right)\\ = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \frac{{{x^2}}}{2}} \right)\\ = 0\end{array}\)

There is a hole at \(\left( {0,0} \right)\).

Find an answer for part (c)

From the graph, it can be observed that there is an inflection point at \(\left( {0.2, - 0.06} \right)\) and a local minimum at \(\left( {0.6, - 0.18} \right)\).

Differentiate the function \(f\left( x \right) = {x^2}\ln x\).

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

If \(f'\left( x \right) > 0\), then;

\(\begin{array}{c}\ln x > - \frac{1}{2}\\x > {e^{ - \frac{1}{2}}}\end{array}\)

So, f is increasing on \(\left( {\frac{1}{{\sqrt e }},\infty } \right)\) and decreasing on \(\left( {0,\frac{1}{{\sqrt e }}} \right)\). By the first derivative test, the local minimum value of the function is \(f\left( {\frac{1}{{\sqrt e }}} \right) = \frac{1}{{2e}}\). The point is approximate \(\left( { - 0.6065,0.1839} \right)\).

Differentiate the equation \(f'\left( x \right) = x\left( {1 + 2\ln x} \right)\).

\(\begin{array}{c}f''\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x\left( {1 + 2\ln x} \right)} \right)\\ = x\left( {0 + \frac{2}{x}} \right) + \left( {1 + 2\ln x} \right)\\ = 3 + 2\ln x\end{array}\)

If \(f''\left( x \right) > 0\), then;

\(\begin{array}{c}\ln x > - \frac{3}{2}\\x > {e^{ - \frac{3}{2}}}\end{array}\)

So, the curve of f is concave upward on \(\left( {{e^{ - \frac{3}{2}}},\infty } \right)\) and concave downward on \(\left( {0,{e^{ - \frac{3}{2}}}} \right)\). The inflection point is \(\left( {{e^{ - \frac{3}{2}}},\frac{{ - 3}}{{2{e^3}}}} \right)\).