# Function domain

• Jul 26th 2005, 07:22 AM
totalnewbie
Function domain
The function is f(x)=-ln(1/x).
1)Find domain.
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.
• Jul 26th 2005, 11:20 PM
ticbol
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.
• Jul 27th 2005, 03:35 AM
totalnewbie
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.

How ?
• Jul 27th 2005, 03:47 AM
ticbol
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?
• Jul 27th 2005, 05:49 AM
totalnewbie
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 ?
• Jul 27th 2005, 11:05 AM
ticbol
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