How do use the first derivative test to determine the local extrema #1/(x^2-x+2)#?

1 Answer
Aug 19, 2015

Answer:

Local maximum: #4/7# (at #x=1/2#.)
There is no local minimum.

Explanation:

#f(x) = 1/(x^2-x+2) = (x^2-x+2)^-1#

#f'(x) = -1(x^2-2+2)^-2 (2x-1)" "# (use power rule and chain rule)

# = -(2x-1)/(x^2-x+2)^2#

#f'(x)# is never undefined (the denominator is never #0#.

#f'(x) = 0# at #x = 1/2#

The only critical number is #1/2#.

Since the denominator of the derivative is always positive, the sign os #f'(x)# is the sanme at that of #-(2x-1)#, which is positive left of #1/2# and negative right of #1/2# so the first derivative test tells us that #f(1/2)# is a local maximum.

#f(1/2) = 1/((1/2)^2-(1/2)+2) = 1/(1/4-1/2+2) = 1/(7/4) = 4/7#