Question

consider the equation below. (if an answer does not exist, enter dne.)

$f(x)=x^{2}-x-ln (x)$

(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)

find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)

(b) find the local minimum and maximum value of $f$.

local minimum value

local maximum value

(c) find the inflection point.

$(x, y)=(quad)$

find the interval(s) on which $f$ is concave up. (enter your answer using interval notation.)

find the interval(s) on which $f$ is concave down. (enter your answer using interval notation.)

Explanation:

Step1: Find the first - derivative

The function is \(f(x)=x^{2}-x - \ln(x)\), \(x>0\).
Using the sum rule \((u + v+w)^\prime=u^\prime + v^\prime+w^\prime\), where \(u = x^{2}\), \(v=-x\), \(w =-\ln(x)\).
The derivative of \(u=x^{2}\) is \(u^\prime = 2x\), the derivative of \(v=-x\) is \(v^\prime=-1\), and the derivative of \(w =-\ln(x)\) is \(w^\prime=-\frac{1}{x}\).
So \(f^\prime(x)=2x - 1-\frac{1}{x}=\frac{2x^{2}-x - 1}{x}=\frac{(2x + 1)(x - 1)}{x}\).

Step2: Determine the intervals of increase and decrease

Set \(f^\prime(x)=0\), then \(\frac{(2x + 1)(x - 1)}{x}=0\) (\(x>0\)).
Since \(2x+1>0\) for \(x>0\), the critical point is \(x = 1\).
Test intervals:
For the interval \((0,1)\), let \(x=\frac{1}{2}\), then \(f^\prime(\frac{1}{2})=\frac{(2\times\frac{1}{2}+1)(\frac{1}{2}-1)}{\frac{1}{2}}=\frac{(1 + 1)(\frac{1}{2}-1)}{\frac{1}{2}}=-2<0\).
For the interval \((1,\infty)\), let \(x = 2\), then \(f^\prime(2)=\frac{(2\times2+1)(2 - 1)}{2}=\frac{5\times1}{2}=\frac{5}{2}>0\).
So \(f(x)\) is decreasing on \((0,1)\) and increasing on \((1,\infty)\).

Step3: Find the local minimum and maximum

Since \(f(x)\) changes from decreasing to increasing at \(x = 1\).
\(f(1)=1^{2}-1-\ln(1)=0\). There is no local maximum because the function only changes from decreasing to increasing (the function has no point where it changes from increasing to decreasing).

Step4: Find the second - derivative

\(f^\prime(x)=2x-1-\frac{1}{x}\), then \(f^{\prime\prime}(x)=2+\frac{1}{x^{2}}\).
Set \(f^{\prime\prime}(x)=0\), \(2+\frac{1}{x^{2}} = 0\) has no solution for \(x>0\) since \(2+\frac{1}{x^{2}}>0\) for all \(x>0\).

Answer:

(a) Increasing: \((1,\infty)\); Decreasing: \((0,1)\)
(b) Local minimum value: \(0\); Local maximum value: DNE
(c) Inflection point: DNE; Concave up: \((0,\infty)\); Concave down: DNE