# How do use the first derivative test to determine the local extrema y= (x²-3x+3)/ (x-1) ?

Jul 30, 2015

There is a local maximum of $- 3$ at $x = 0$
and a local minimum of $1$ at $x = 2$.

#### Explanation:

Let f(x) = (x²-3x+3)/(x-1)

Differentiate using the quotient rule and simplify to get:

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

$f ' \left(x\right) = 0$ at $0$ and $2$ and $f ' \left(x\right)$ is not defined at $1$.

The domain of $f$ includes all Real numbers except $1$,

so the critical numbers for $f$ are: $0$ and $2$

Test the critical number $0$

If $x$ is a little less that $0$, then $x$ and $x - 2$ are negative, while ${\left(x - 1\right)}^{2}$ is positive, so $f ' \left(x\right) > 0$

When $x$ is a little greater than $0$, the sign of $x$ is positive, but the signs of the other factor remain the same. So $f ' \left(x\right)$ changes to a negative value.

This tells us that $f \left(0\right)$ (which is $- 3$) is a local maximum.

Test the critical number $2$

For$x$ a little less than $2$, $f ' \left(x\right) < 0$ and
For $x$ a little greater than $2$, the sign changes to $f ' \left(x\right) > 0$

Therefore $f \left(2\right)$ (which is $\frac{4 - 6 + 3}{1} = 1$) is a local minimum.