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

Question

Christopher Allers · Accepted Answer

Strictly speaking it is neither. #f(1)=0# is a constant.

However, #f'(x) = 2(x-2)(x-1)+(x-2)^2#, so #f'(1)=1#. Note that #f'# is continuous near #1#. So, #f'(x)# is positive near #1# and #f(x)# is increasing near #1#. Note For a number #a#, the phrase "near #a#" means "in some open interval containing #a#"

Paisley Johnson · Answer

undefined To determine whether $ f(x) = (x - 2)^2(x - 1) $ is increasing or decreasing at $ x = 1 $, we can analyze the sign of the derivative $ f'(x) $ at that point.

First, let's find the derivative $ f'(x) $:

$$ f(x) = (x - 2)^2(x - 1) $$

$$ f'(x) = 2(x - 2)(x - 1) + (x - 2)^2 \cdot 1 $$

Now, plug in $ x = 1 $ into $ f'(x) $ to evaluate the sign of the derivative at $ x = 1 $:

$$ f'(1) = 2(1 - 2)(1 - 1) + (1 - 2)^2 \cdot 1 $$

$$ f'(1) = 2(-1)(0) + (-1)^2 \cdot 1 $$

$$ f'(1) = 1 $$

Since $ f'(1) = 1 $, which is positive, the function $ f(x) $ is increasing at $ x = 1 $.

Aurora Alvarado · Answer

undefined To determine if the function $ f(x) = (x - 2)^2(x - 1) $ is increasing or decreasing at $ x = 1 $, we need to examine the sign of the derivative of the function at $ x = 1 $.

To find the derivative of $ f(x) $, we can use the product rule and the chain rule. After finding the derivative, we evaluate it at $ x = 1 $ to determine if it is positive (indicating the function is increasing) or negative (indicating the function is decreasing) at that point.