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

Aug 19, 2015

Local maximum: $\frac{4}{7}$ (at $x = \frac{1}{2}$.)
There is no local minimum.

#### Explanation:

$f \left(x\right) = \frac{1}{{x}^{2} - x + 2} = {\left({x}^{2} - x + 2\right)}^{-} 1$

$f ' \left(x\right) = - 1 {\left({x}^{2} - 2 + 2\right)}^{-} 2 \left(2 x - 1\right) \text{ }$ (use power rule and chain rule)

$= - \frac{2 x - 1}{{x}^{2} - x + 2} ^ 2$

$f ' \left(x\right)$ is never undefined (the denominator is never $0$.

$f ' \left(x\right) = 0$ at $x = \frac{1}{2}$

The only critical number is $\frac{1}{2}$.

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

$f \left(\frac{1}{2}\right) = \frac{1}{{\left(\frac{1}{2}\right)}^{2} - \left(\frac{1}{2}\right) + 2} = \frac{1}{\frac{1}{4} - \frac{1}{2} + 2} = \frac{1}{\frac{7}{4}} = \frac{4}{7}$