Question

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

Answer 1

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`