Question

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

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

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

Short answer

(a) The function \(f(x)\) increasing on \((1, + \infty )\) and decreasing on \((0,1)\).

(b) There is no local maximum and the local minimum is \(f(1) = 0\).

(c) The function \(f(x)\) is concave upward on \((0,\infty )\) and there is no inflection point.

Step-by-step solution

Given data

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

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:

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

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

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

Hence, the critical points occurs at \(x = \frac{{ - 1}}{2},\;x = 1\) and in the interval notation\((0,\;1) \cup (1,\;\infty )\).

Determine the nature of \(f'(x)\)

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

Interval	\(f'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( {x - 1} \right)}}{x}\)	\(f\left( x \right)\)
\(0 < x < 1\)	\(\begin{aligned}{c}f'\left( {\frac{1}{2}} \right) &= \frac{{\left( {2\left( {\frac{1}{2}} \right) + 1} \right)\left( {\frac{1}{2} - 1} \right)}}{{\frac{1}{2}}}\\ &= \frac{{\left( {1 + 1} \right)\left( { - \frac{1}{2}} \right)}}{{\frac{1}{2}}}\\ &= - 2\\f'\left( x \right) < 0\end{aligned}\)	Decreasing on \(\left( {0,1} \right)\)
\(1 < x < \infty \)	\(\begin{aligned}{c}f'\left( 2 \right) &= \frac{{\left( {2\left( 2 \right) + 1} \right)\left( {2 - 1} \right)}}{2}\\ &= \frac{{\left( {4 + 1} \right)\left( 1 \right)}}{2}\\ &= \frac{5}{2}\\f'\left( x \right) > 0\end{aligned}\)	Increasing on \(\left( {1,\infty } \right)\).

Table 1

From the above table, it is observed that, \(f(x)\) is increasing on \((1,\infty )\) and decreasing on\((0,1)\).

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

From part (a), the function \(f(x)\) is decreasing at \(x = 1\)

Thus, there is no local maximum and the local minimum occurs at \(x = 1\)

Substitute \(x = 1\) in \(f(x)\) and obtain the local minimum as follows:

\(\begin{aligned}{c}f(1) &= {(1)^2} - (1) - \ln (1)\\ &= 1 - 1 - 0\\ &= 0\end{aligned}\)

Thus, there is no local maximum and local minimum is \(f(1) = 0\).

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

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

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{d}{{dx}}\left( {{f^\prime }(x)} \right)\\ &= \frac{d}{{dx}}\left( {2x - 1 - \frac{1}{x}} \right)\\ &= 2(1) - 0 - \left( {\frac{{ - 1}}{{{x^2}}}} \right)\\ &= 2 + \frac{1}{{{{\rm{x}}^2}}}\end{aligned}\)

Since \({f^{\prime \prime }}(x) = 2 + \frac{1}{{{x^2}}}\) is always positive, \(f\) is concave upward in the domain \((0,\infty )\).

Definition used: Inflection point

"A point\(P\)on a curve under the function\(f(x)\)is called an inflection point if\(f\)is continuous and curve changes from concave upward to concave downward or from concave downward to concave upward at\(P\).”

According to the definition the given function has no inflection point.