Please help me understand this. Someone already helped me find the derivative of
ID: 2866499 • Letter: P
Question
Please help me understand this. Someone already helped me find the derivative of y= x-4ln (3x-9) which is (x-7)/(x-3). Now I'm supposed to find the critical points and the Intervals of increase and decrease. When I look at a graph of the original equation it makes sense but that won't help me on a test. For example, how am I supposed to know There are no points on the graph before x=3. Am I supposed to plot any possible asymptotes? Also if the denominator is x-3, why isn't 3 a critical point. Thanks to anyone who can explain this to me.
Explanation / Answer
Yes, your derivative is spot on
Now, the function f(x) = x - 4*ln(3x - 9), which is a LOG function
You are supposed to know that the argument within a Log or Ln must ALWAYS be positive
So, in this case, the argument was 3x - 9
So, the condition for x, the domain becomes :
3x - 9 > 0
3x > 9
x > 3 ----> This is how there are no points before x = 3
Also, the function is not defined for x = 3 itself
The derivative is (x - 7) / (x - 3)
So, when we equate derivative to 0, we get :
(x - 7) / (x - 3) = 0
x - 7 = 0
x = 7 ---> This is the ONLY CRITICAL POINT
The denominator x = 3 is NOT a critical number because the function itself is not defined for x = 3.
It is defined ONLY for x > 3 as explained above.....
So, x = 7 being the only critical points, we know that the intervals of interest are :
(3 , 7) and (7 , infinity)
Region 1 : (3 , 7)
Lets choose testvalue = 4 say
f'(x) = (x - 7) / (x - 3)
So, f'(4) = (4 - 7) / (4 - 3) --> -3/1 --> -3 ---> negative
So, since the derivative is NEGATIVE, we know that the function DECREASES over (3 , 7)
Region 2 : (7 , inf)
Lets choose testvalue = 8 say
f'(8) = (8 - 7) /(8 - 3)
f'(8) = 1/5 ---> positive
So, the function INCREASES over (7 , inf)
S, your intervals of inc and dec will be written as :
Increase : (7 , inf)
Decrease : (3 , 7)
I hope this cleared things :)