Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x^
ID: 2861983 • Letter: C
Question
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x^2 - x - ln(x) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) Find the local minimum and maximum value of f. Find the inflection point. Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)Explanation / Answer
ln x is valid only when x>=0 so keep this in mind
a) increasing if fdash(x) >0
f dash(x) = 2x-1-1/x
2x-1-1/x > 0, solving we get -1/2 <x <0 and x>1
but we cannot have x<0 so only x>1 is the answer
b) decreasing is f dash(x) <0
2x-1-1/x < 0, solving we get 0 <x <1 and x< -1/2
we cannot have x<0 so only 0<x<1 is the answer
c)
f double dash (x) = 2 +1/x^2
for max/min , 2+1/x^2 = 0 => x^2 = -1/2 => not possible so no max value
for local minima , we need to have x = 0 is the starting point in the range so
f(0) = 0 so 0 is the minimum value
concave up if 2+1/x^2 > 0 => solution is x>0
concave down => 2+1/x^2 < 0 => no solution as x^2 term is always positive