Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x2

Question

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x2-x-In() (a) Find the interval on which fis increasing. (Enter your answer using interval notation.) Find the interval on which fis 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. Find the interval on which fis concave up. (Enter your answer using interval notation.) Find the interval on which fis concave down. (Enter your answer using interval notation) Need Help? Read It Talk to a Tutor

Explanation / Answer

f(x) = x2-x -ln x

f '(x) = 2x -1 -(1/x)

Critical numbers, f '(x) =0

2x-1-(1/x)=0

2x2-x-1=0

(2x+1)(x-1)=0

x=-1/2 , 1

In (-inf,-1/2) , f'(x)<0

In (-1/2, 1), f'(x)<0

In (1,inf), f'(x)>0

a. Increasing interval (1,inf)

b. decreasing interval (0,1)

f''(x) = 2+(1/x2)

Inflection point

f ''(x)=0

2+(1/x2)=0

No real values

No inflection point

f ''(-1/2) = 6>0

f''(1)= 2+1 =3 >0

local minima (1,0)

No local maximum

Concave up (0,inf)

concave down - none

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