Is #f(x)=(x-2)^2/(x-1)# increasing or decreasing at #x=2#?

Question

Christopher Agresta · Accepted Answer

Find the first derivative , then determine if it is positive (increasing) or negative (decreasing) at #x=2#

Using the quotient rule , then simplify ...

#f'=((x-2) x)/(x-1)^2#

Now, #f'(2)=0#

So, at #x=2#, the derivative equals zero which indicates a critical point that is neither increasing or decreasing .

Hope that helped

Adam Adkins · Answer

undefined To determine whether $ f(x) = \frac{{(x - 2)^2}}{{x - 1}} $ is increasing or decreasing at $ x = 2 $, we need to analyze the sign of the derivative of the function at that point.

First, find the derivative of $ f(x) $:
$$ f'(x) = \frac{{2(x - 2)(x - 1) - (x - 2)^2}}{{(x - 1)^2}} $$

Now, evaluate $ f'(2) $ to determine the behavior of the function at $ x = 2 $:
$$ f'(2) = \frac{{2(2 - 2)(2 - 1) - (2 - 2)^2}}{{(2 - 1)^2}} $$
$$ f'(2) = \frac{{2(0)(1) - 0}}{{1}} $$
$$ f'(2) = 0 $$

Since the derivative $ f'(2) $ is zero, we cannot determine whether the function is increasing or decreasing at $ x = 2 $ using the first derivative test. Additional analysis may be required, such as considering the behavior of the function around the point $ x = 2 $.