Question

(a) Determine the intervals on which the function \(f(x) = {x^2}\ln x\) is increasing or decreasing.

(b) Determine the local maximum and minimum values of \(f(x) = {x^2}\ln x\).

(c) Determine the intervals of concavity and the inflection points of \(f(x) = {x^2}\ln x\).

Short answer

(a) The function \(f(x)\) is increasing on \(\left( {\frac{1}{{\sqrt e }},\infty } \right)\) and is decreasing on. \(\left( {0,\frac{1}{{\sqrt e }}} \right)\).

(b) There is no local maximum and the local minimum is, \(f\left( {\frac{1}{{\sqrt e }}} \right) = \frac{{ - 1}}{{2e}}\).

(c) The \(f(x)\) is concave upward on \(\left( {{e^{\frac{{ - 3}}{2}}}, + \infty } \right)\) and concave downward on \(\left( {0,{e^{\frac{{ - 3}}{2}}}} \right)\) and the inflection point is \(\left( {{e^{\frac{{ - 3}}{2}}}, - 0.0747} \right)\).

Step-by-step solution

Given data

The given function is, \(f(x) = {x^2}\ln x\).

Concept of Increasing/decreasing test and concavity

Increasing/decreasing test:

(a) If\({f^\prime }(x) > 0\)on an interval, then\(f\)is increasing on that interval.

(b) If\({f^\prime }(x) < 0\)on an interval, then\(f\)is decreasing on that interval.

Definition of concavity:

"(a) If\({f^\prime }(x) > 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave upward on\(I\).

(b) If \({f^\prime }(x) < 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave downward on\(I\).”

(a)Step 3: Obtain the critical points

Obtain the derivative of \(f(x)\) as follows:

\({f^\prime }(x) = \frac{d}{{dx}}\left( {{x^2}\ln x} \right)\)

Apply the Product Rule as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= {x^2}{(\ln x)^\prime } + (\ln x){\left( {{x^2}} \right)^\prime }\\ &= {x^2}\left( {\frac{1}{x}} \right) + \ln x(2x)\\ &= x(1 + 2\ln x)\end{aligned}\)

Set \({f^\prime }(x) = 0\) and obtain the critical points as follows:

\(\begin{aligned}{c}x(1 + 2\ln x) &= 0\\x = 0,\;1 + 2\ln x &= 0\\x = 0,\;\ln x &= \frac{{ - 1}}{2}\\x = 0,\;x &= {e^{\frac{{ - 1}}{2}}}\end{aligned}\)

Hence, the critical points occurs at \(x = 0,\;x = {e^{\frac{{ - 1}}{2}}}\left( {x = \frac{1}{{\sqrt e }} \approx 0.6065} \right)\) and in the interval notation \(\left( {0,\;\frac{1}{{\sqrt e }}} \right) \cup \left( {\frac{1}{{\sqrt e }},\;\infty } \right).\)

Determine the nature of \(f'(x)\)

Determine the nature of \({f^\prime }(x)\) in table 1 as follows:

Interval	\(f'\left( x \right) = x\left( {1 + 2\ln x} \right)\)	\(f\left( x \right)\)
\(0 < x < \frac{1}{{\sqrt e }}\)	\(\begin{aligned}{c}f'\left( {\frac{1}{2}} \right) &= \frac{1}{2}\left( {1 + 2\ln \frac{1}{2}} \right)\\ &= \frac{1}{2}\left( {1 + 2\left( { - 0.693} \right)} \right)\\ &= - ve\\f'\left( x \right) < 0\end{aligned}\)	Decreasing on \(\left( {0,\frac{1}{{\sqrt e }}} \right)\)
\(\left( {{e^{\frac{{ - 3}}{2}}}, + \infty } \right) \cup \left( {0,{e^{\frac{{ - 3}}{2}}}} \right)\)	\(\begin{aligned}{c}f'\left( 1 \right) &= 1\left( {1 + 2\ln 1} \right)\\ &= \frac{1}{2}\left( {1 + 2\left( 0 \right)} \right)\\ &= + ve\\f'\left( x \right) > 0\end{aligned}\)	Increasing on \(\left( {\frac{1}{{\sqrt e }},\infty } \right)\)

From the above table observe that, \(f(x)\) is increasing on \(\left( {\frac{1}{{\sqrt e }}, + \infty } \right)\) and is decreasing on \(\left( {0,\frac{1}{{\sqrt e }}} \right)\).

(b)Step 5: Determine the local maximum and local minimum

From part (a), the function \(f(x)\) is decreasing at \(x = \frac{1}{{\sqrt e }}\).

Thus, there is no local maximum and the local minimum occurs at \(x = \frac{1}{{\sqrt e }}\).

Substitute \(x = \frac{1}{{\sqrt e }}\) in \(f(x)\) and obtain the local minimum as follows:

\(\begin{aligned}{c}f\left( {\frac{1}{{\sqrt e }}} \right) &= {\left( {\frac{1}{{\sqrt e }}} \right)^2}\ln \left( {\frac{1}{{\sqrt e }}} \right)\\ &= \frac{1}{e}\left( {\frac{{ - 1}}{2}} \right)\\ &= \frac{{ - 1}}{{2e}}\end{aligned}\)

Thus, there is no local maximum and local minimum is \(f\left( {\frac{1}{{\sqrt e }}} \right) = \frac{{ - 1}}{{2e}}\).

(c)Step 6: Obtain the derivative of \({f^{\prime \prime }}(x)\)

Obtain the derivative of \({f^{\prime \prime }}(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{d}{{dx}}\left( {{f^\prime }(x)} \right)\\ &= \frac{d}{{dx}}(x + 2x\ln x)\\ &= \frac{d}{{dx}}(x) + \frac{d}{{dx}}(2x\ln x)\\ &= \frac{d}{{dx}}(x) + \left( {2x{{(\ln x)}^\prime } + \ln x{{(2x)}^\prime }} \right)\end{aligned}\)

Simplify further to obtain \({f^{\prime \prime }}(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= 1 + \left( {2x \cdot \left( {\frac{1}{x}} \right) + \ln x(2)} \right)\\ &= 1 + (2 + 2\ln x)\\ &= 3 + 2\ln x\end{aligned}\)

Set \({f^{\prime \prime }}(x) = 0\) and solve for \(x\)as follows:

\(\begin{aligned}{c}3 + 2\ln x &= 0\\\ln x &= \frac{{ - 3}}{2}\\x &= {e^{\frac{{ - 3}}{2}}}\end{aligned}\)

In the interval notation \(\left( {0,\,{e^{\frac{{ - 3}}{2}}}} \right)\).

Determine the nature of \({f^{\prime \prime }}(x)\)

Determine the nature of \({f^{\prime \prime }}(x)\) in table 2 as follows:

Interval	\(f''\left( x \right) = 3 + 2\ln x\)	\(f\left( x \right)\)
\({e^{ - \frac{3}{2}}} < x < \infty \)	\(\begin{aligned}{c}f''\left( 1 \right) &= 3 + 2\ln \left( 1 \right)\\ &= 3\\ &= + ve\\f''\left( x \right) > 0\end{aligned}\)	Concave upward on \(\left( {{e^{ - \frac{3}{2}}},\infty } \right)\)
\(0 < x < {e^{ - \frac{3}{2}}}\)	\(\begin{aligned}{c}f''\left( {0.1} \right) &= 3 + 2\ln \left( {0.1} \right)\\ &= 3 + 2\left( { - 2.302} \right)\\ &= - ve\\f''\left( x \right) < 0\end{aligned}\)	Concave downward on \(\left( {0,{e^{ - \frac{3}{2}}}} \right)\)

Table 2

Hence, \(f(x)\) is concave upward on \(\left( {{e^{\frac{{ - 3}}{2}}}, + \infty } \right)\) and concave downward on \(\left( {0,{e^{\frac{{ - 3}}{2}}}} \right)\). Here, the curve changes from concave downward to concave upward at \(x = {e^{\frac{{ - 3}}{2}}}\).

Find inflection point

Substitute \(x = {e^{\frac{{ - 3}}{2}}}\) in \(f(x)\) and obtain the inflection point as follows:

\(\begin{aligned}{c}f\left( {{e^{\frac{{ - 3}}{2}}}} \right) &= {\left( {{e^{\frac{{ - 3}}{2}}}} \right)^2}\ln \left( {{e^{\frac{{ - 3}}{2}}}} \right)\\ &\approx - 0.0747\end{aligned}\)

Thus, the inflection point is \(\left( {{e^{\frac{{ - 3}}{2}}}, - 0.0747} \right)\).