consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{2}-x-ln (x)$\n\n(a) find…

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