Asked by Joshua
Hello, my question is the following:
The derivative of f(x) = ln(x) is given by f'(x) = 1/x. Find an equation of the tangent line to the graph y = ln(x) that passes through the point (0,0).
So if I go through this using the typical method I use I have the point slope formula being y- y1 = m(x - x1).
This gives me y - 0 = m(x - 0) given the values (0,0).
My m is provided which is f'(x) = 1/x.
so i get y - 0 = (1/x)(x-0)
At this point I realize y = (1/x)(x) is what is left over and that is not giving me the correct answer as it would appear to give me 1.
Am I using the wrong method here or am I making a mistake along the way? Any help is appreciated.
The derivative of f(x) = ln(x) is given by f'(x) = 1/x. Find an equation of the tangent line to the graph y = ln(x) that passes through the point (0,0).
So if I go through this using the typical method I use I have the point slope formula being y- y1 = m(x - x1).
This gives me y - 0 = m(x - 0) given the values (0,0).
My m is provided which is f'(x) = 1/x.
so i get y - 0 = (1/x)(x-0)
At this point I realize y = (1/x)(x) is what is left over and that is not giving me the correct answer as it would appear to give me 1.
Am I using the wrong method here or am I making a mistake along the way? Any help is appreciated.
Answers
Answered by
Reiny
the whole problem becomes trivial when you realize that (0,0) does not lie on y = lnx
If you recall your properties and definition of logs, you know that we can only take logs of positive numbers, that is, ln (0) is undefined.
Your method is correct until you come to the point, where the slope would be 1/0 , which of course is undefined.
To make a long story short, the whole question is bogus, since the given point does lie on your curve, thus no tangent is possible.
If you recall your properties and definition of logs, you know that we can only take logs of positive numbers, that is, ln (0) is undefined.
Your method is correct until you come to the point, where the slope would be 1/0 , which of course is undefined.
To make a long story short, the whole question is bogus, since the given point does lie on your curve, thus no tangent is possible.
Answered by
Steve
I think you both misread the question. We want a line tangent to ln(x) which passes through (0,0).
Since the slope of the tangent is 1/x, we want a line through (0,0) and (h,ln(h)) with slope 1/h.
Clearly, at x=e, y=1 and the slope is 1/e, so
y = x/e is the tangent at (e,1)
See
http://www.wolframalpha.com/input/?i=plot+y%3Dln%28x%29%2C+y%3Dx%2Fe+for+x+%3D+0..10
Since the slope of the tangent is 1/x, we want a line through (0,0) and (h,ln(h)) with slope 1/h.
Clearly, at x=e, y=1 and the slope is 1/e, so
y = x/e is the tangent at (e,1)
See
http://www.wolframalpha.com/input/?i=plot+y%3Dln%28x%29%2C+y%3Dx%2Fe+for+x+%3D+0..10
Answered by
Joshua
My professor claims that it is solvable because it says that the point (0,0) lieS on a tangent line to the graph y=ln(x) at some point (a,b)
Answered by
Joshua
ah thank you steve
Answered by
Joshua
why does the line have to cross (h, ln(h))?
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