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

Question

Christopher Allers · Accepted Answer

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

#f'(x) = ((2x-4)(x-3)-(x^2-4x+5)(1))/(x-3)^2# # = (x^2-6x+7)/(x-3)^2# #f'# fails to exists only at #x=3# which is not in the domain of #f#, and #f'(x) = 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.

Cameron Alexander · Answer

undefined To find the critical points of $ f(x) = \frac{{x^2 - 4x + 5}}{{x - 3}} $, you need to first find its derivative, set it equal to zero, and solve for $ x $. Then, check for any potential vertical asymptotes.

1. Find the derivative of $ f(x) $ using the quotient rule:
$$ f'(x) = \frac{{(x - 3)(2x - 4) - (x^2 - 4x + 5)(1)}}{{(x - 3)^2}} $$

2. Simplify $ f'(x) $:
$$ f'(x) = \frac{{2x^2 - 10x + 12 - x^2 + 4x - 5}}{{(x - 3)^2}} $$
$$ f'(x) = \frac{{x^2 - 6x + 7}}{{(x - 3)^2}} $$

3. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ x^2 - 6x + 7 = 0 $$
$$ (x - 1)(x - 7) = 0 $$
$$ x = 1 \quad \text{or} \quad x = 7 $$

4. Check for vertical asymptotes by identifying any values of $ x $ that make the denominator of $ f(x) $ equal to zero:
$$ x - 3 = 0 $$
$$ x = 3 $$

So, the critical points of $ f(x) = \frac{{x^2 - 4x + 5}}{{x - 3}} $ are $ x = 1 $ and $ x = 7 $, and there is a potential vertical asymptote at $ x = 3 $.