
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.

Domain is the set of xvalues 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.

I encountered an another problem on my function.
It is said to make the function y=f(x) tangent of the graph where xcoorinate is 1.
I did it and and the equation of tangent is: y=x1
But the following point said to make square function g(x)=ax^2+c according to the circumstance that tangent y=x1 would be tangent of the graph y=g(x) whose xcoordinate is 1.
How ?

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=x1"
That alone will confuse anyone. Because, how did you get that?

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 xcoordinate which was 1 I found the slope which was 1 too, because f'(xcoordinate)=slope.
General equation of tangent is the following: yy0=k(xx0) where k is the slope. So I replaced the numbers: y+y0=1*(x1) where y0=y(x0)=ln(1)=0
And then y+0=1*(x1) which is actually y=x1. 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=x1 would be the tangent of graph y=g(x) whose xcoordinate is 1.
Did you understand my english ?

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 = x1
>>>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(xx0)" should be
yy0=k(xx0) 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 = x1?
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.

Yes, your understanding was really correct. g(x) = (1/2)x^2 1/2 is the right answer. Thank you.

You post it separately and I will answer it.
One posting, one question. Or, new question, new posting.