What are the intervals for which the derivative of f(x) is positive or negative?

[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 don't get the right answer.

can anyone help?:yuck:

 

[VietDao29]

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 don't get the right answer.

can anyone help?:yuck:

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? :)

 

[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 don't 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 don't know how to articulate it as well.

 

[hellraiser]

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

 

[HallsofIvy]

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 don't know if you can do this across an inequality):

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?)

 

[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