Find the intervals on which x^2/(x-3)^2 is decreasing.
a. (0,6)
b. (0,3), (3,6)
c. (-inf,0), (6,inf)
The critical numbers are x=0 and 6, but since there is an asymptote at x=3, I'm not sure whether a or b is the answer.
a function is decreasing if the first derivative is negative.
It also helps to know what the graph looks like
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2%2F(x-3)%5E2
Where does the critical value of x = 6 come from?
I'm sorry...the function is supposed to be x^2/(x-3).
To determine the intervals on which the function x^2/(x-3)^2 is decreasing, we need to analyze the behavior of its derivative.
First, let's find the derivative of the function:
f(x) = x^2/(x-3)^2
To find f'(x), we can use the quotient rule:
f'(x) = [(x-3)^2(2x) - (x^2)(2(x-3))]/(x-3)^4
= [2x(x-3)^2 - 2x^2(x-3)]/(x-3)^4
= [2x(x-3)(x-3 - x)]/(x-3)^4
= -2x/(x-3)^3
Next, we need to determine where the derivative is negative (indicating a decreasing interval).
Since we have a denominator of (x-3)^3, it indicates a vertical asymptote at x = 3. Therefore, the function is not defined at x = 3.
Now, let's analyze the sign of the derivative on intervals around the critical numbers:
1. For x < 3, choose x = 0:
f'(0) = -2(0)/(0-3)^3 = 0
The derivative is zero (neither positive nor negative) at x = 0.
2. For x > 3, choose x = 6:
f'(6) = -2(6)/(6-3)^3 = -2/27
The derivative is negative at x = 6.
From this analysis, we can conclude that the function is decreasing on the interval (0, 6). Therefore, the correct option is a. (0, 6).
Option b. (0, 3) and (3, 6) is not correct because the function is not defined at x = 3 and does not have a decreasing behavior in the interval (3, 6).
Option c. (-inf, 0) and (6, inf) is not correct because the function is decreasing only on the interval (0, 6), not on the intervals (-inf, 0) and (6, inf).
So, the answer is a. (0, 6).