Answer:
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
