Differentiating natural logs

1. Feb 2, 2006

QueenFisher

let f(x) = ln(x-2)+ln(x-6). Write down the natural domain of f(x).

i got this bit right, it's x>6

find f'(x)

i got 1/(x-2) + 1/(x-6)

which i think is right.

then it says find the intervals for which f'(x) is a. positive and then b. negative.

for a. i put f'(x)>0 and solved for x and i got the right answer, but i'm not sure if this is the right method.

because for b. if i put f'(x)<0 and solve for x i dont get the right answer.

can anyone help???:yuck:

2. Feb 2, 2006

VietDao29

May I ask you to show your step, so that we can verify it for you?
Since you do know how to start the problem, you are getting wrong results, it's because either there's a typo in the book, or you have done something wrong.
So can you show your work? :)

3. Feb 2, 2006

QueenFisher

ok then:
for a.

1/(x-2) + 1/(x-6)>0
so
1/(x-2)>-1/(x-6)

cross multiplying (thi is the bit i'm not sure about - i dont know if you can do this across an inequality):

x-6>-1(x-2)
gives
x>4

a similar process for part b. gives x<4 but the textbook says there are no values for which f'(x) is negative.

i know about the whole log graph shape thingy, and that if the logs are positive the gradient is positive always but i'm not too sure about this really and i dont know how to articulate it as well.

4. Feb 2, 2006

hellraiser

You got the domain x>6. And if f'(x) is to be negative, x<4. That is a contradiction.

5. Feb 2, 2006

HallsofIvy

Staff Emeritus
That's very likely your problem- Multiplying on both sides of inequality by a negative number changes the direction of the inequality. If you multiply both sides by something involving the variable, you don't know whether it is positive or negative! In this case, it should be obvious that if x> 6 (which you have already said is the domain of this function) then both x-2 and x- 6 are positive so both fractions are positive. The sum of two positive numbers is positive. The derivative, 1/(x-2)+ 1/(x- 6), is positive for all x> 6 which is the domain of the function. The derivative is always positive, never negative on the domain of the function.

(Why are so many people posting derivative, integral, and even differential equations under precalculus??? Has the definition of "calculus" been changed and I didn't notice?)

Last edited: Feb 2, 2006
6. Feb 3, 2006

QueenFisher

everything in that forum looks really hard! and anyway i wouldn't know which forum to post in if i had a different question cos my maths isn't structured into precalculus and the like so i just post here