Consider the equation below. (if an answer does not exist, enter dne.)f(x) = x2 − x − ln(x)

(a) find the interval on which f is increasing. find the interval on which f is decreasing.

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

(c) find the inflection point.(x, y) = find the interval on which f is concave up. find the interval on which f is concave down.

a) decreasing from (0,2),  increasing from (2,∞ )

b) local minimum in x=2 . there is no maximum or minimum value

c) DNE. there is no inflexion point

Step-by-step explanation:

f(x) = x² - x - ln (x)

since ln(x) is defined for positive values only x must be greater than 0 (x>0)

also we will need the first derivative and the second derivative with respecto to x

f(x) = x² - x - ln (x)

df/dx (x) = 2x - 1 - 1/x

d²f /(dx)² (x) = 2 + 1/x²

a) to find the increasing and decreasing intervals we will need to evaluate the rate of change (df/dx) :

df/dx = 0 when 2x - 1 - 1/x = 0 →  2x² - x - 1 = 0   → x = (1±√(1+8))/2 = (1 ± 3)/2

→ x1 = 2 , x2 = -1 (discarded because x2<0)

therefore since 2x increases and 1/x decreases with increasing x

for x > 2  , df/dx is positive and thus f increases with increasing x

for 0<x< 2,  df/dx is negative and thus f decreases with increasing x

b) since f increases with increasing x for x> 2 and decreases with increasing x for, 0<x< 2 , f should be a minimum value.

we can verify it with the second derivative

d²f /(dx)² (x) =2 + 1/x² → for x >0 , d²f /(dx)² is always >0 therefore

d²f /(dx)² (x1) > 0 and df/dx (x1) =0 → thus f(x) is a local minimum of x

there are no maximum values since for x → ∞ , f(x) → ∞ and for x→ 0 → f(x) → -∞ (because of the ln(x) function)

c) there are no inflexion points since  d²f /(dx)² (x1) is always greater than 0 for x>0

your answer is a

step-by-step explanation:

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