The function is f(x)=-ln(1/x).
Denominator must not be zero and the whole value as well. If the domain is [1;infinity) I reckon it is wrong because when the denominator is for example 9 milliard the answer is not calculated becuase it's value is negative.
July 26th 2005, 11:20 PM
Domain is the set of x-values that will define or validate f(x), or make f(x) true.
ln is natural log.
(-infinity,0] don't define logs, so x cannot be from -infinity up to zero.
(0,infinity) will define logs, so x can be from, but not including, 0 to infinity.
f(x) = -ln(1/x)
f(x) = (-1)ln(1/x)
f(x) = ln[(1/x)^(-1)]
f(x) = ln(1 / 1/x)
f(x) = ln(x)
That means x can be from right after zero up to right before infinity.
Therefore, domain of f(x) is (0,infinity). ----answer.
July 27th 2005, 03:35 AM
I encountered an another problem on my function.
It is said to make the function y=f(x) tangent of the graph where x-coorinate is 1.
I did it and and the equation of tangent is: y=x-1
But the following point said to make square function g(x)=ax^2+c according to the circumstance that tangent y=x-1 would be tangent of the graph y=g(x) whose x-coordinate is 1.
July 27th 2005, 03:47 AM
Don't understand what you mean.
Why don't you post the original question as you see it from the source. I mean, do not interpret the question. Copy it as it is.
"I did it and and the equation of tangent is: y=x-1"
That alone will confuse anyone. Because, how did you get that?
July 27th 2005, 05:49 AM
I have to interpret the question because the questions were not in english.
The function was f(x) = ln(x) and it's derivative is y=1/x. According to the x-coordinate which was 1 I found the slope which was 1 too, because f'(x-coordinate)=slope.
General equation of tangent is the following: y-y0=k(x-x0) where k is the slope. So I replaced the numbers: y+y0=1*(x-1) where y0=y(x0)=ln(1)=0
And then y+0=1*(x-1) which is actually y=x-1. And this way I got this equation of tangent
The equation of tangent was easy but the next is complicated.
How to determine letters a and c on the square function g(x)=ax^2+c according to the circumstance that tangent y=x-1 would be the tangent of graph y=g(x) whose x-coordinate is 1.
Did you understand my english ?
July 27th 2005, 11:05 AM
I understand your English. That is not the problem. The problem is I don't understand very well what you mean.
So you have a function
f(x) = ln(x) ----(i)
You were told to find the tangent line to (i) at the point where x=1.
And you found:
>>>slope of the tangent line there is 1.
>>>equation of tangent line is y = x-1
>>>the point on f(x) where x=1 is (1,0)
Very good, those are all correct.
Although "General equation of tangent is the following: y+y0=k(x-x0)" should be
y-y0=k(x-x0) -----not y+y0.
Now you are told to find a new function such that at point (1,0) of the graph of this new function the tangent line is the same y = x-1?
And this new function is
g(x) = ax^2 +c ? ---------(ii)
If my understanding then is correct,
We get the slope of the tangent line (the slope is 1, allright),
Differentiate both sides of (ii) with respect to x,
g'(x) = 2ax
Since that should be equal to 1, and x=1, then,
1 = 2a(1)
a = 1/2 ----***
Substitute those into (ii), knowing g(x) = 0 when x=1, point (1,0) remember?,
0 = (1/2)(1^2) +c
0 = 1/2 +c
c = -1/2 ----***
Therefore, the new function is
g(x) = (1/2)x^2 -1/2 ------------answer.
July 27th 2005, 12:56 PM
Yes, your understanding was really correct. g(x) = (1/2)x^2 -1/2 is the right answer. Thank you.
July 27th 2005, 09:56 PM
You post it separately and I will answer it.
One posting, one question. Or, new question, new posting.