# What are the critical points of f(x) =(x^2-4x+5)/(x-3)?

Feb 4, 2016

The critical numbers are $1$ and $7$.

#### Explanation:

$f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(x - 3\right) - \left({x}^{2} - 4 x + 5\right) \left(1\right)}{x - 3} ^ 2$

$= \frac{{x}^{2} - 6 x + 7}{x - 3} ^ 2$

$f '$ fails to exists only at $x = 3$ which is not in the domain of $f$,

and $f ' \left(x\right) = 0$ at $x = 1$ and at $x = 7$, both of which are in the domain of $f$.

So, the critical numbers for $f$ are $1$ and $7$.

There appears to be some variability in the use of the term "critical point".
I use it to mean a point in the domain of a function at which the derivative is $0$ or fails to exist.
Some people seem to use it to mean a point on the graph of a function at which the derivative is $0$ or fails to exist.