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 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.
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
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)
ah thank you steve
why does the line have to cross (h, ln(h))?
You are on the right track, but there seems to be a slight error in your calculations. Let's go through the process step-by-step to find the equation of the tangent line.
First, we know that the point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
In this case, we are given that the tangent line passes through the point (0, 0). So, we can write the equation of the tangent line as y - 0 = m(x - 0).
Now, we need to find the slope, which is given by the derivative of f(x). As you correctly mentioned, the derivative of f(x) = ln(x) is f'(x) = 1/x.
Since the tangent line passes through the point (0, 0), we need to find the slope at x = 0. However, the derivative 1/x is undefined at x = 0. Therefore, we need to use a limit to find the slope at x = 0.
Taking the limit of f'(x) as x approaches 0:
lim x->0 (1/x) = ∞
The slope of the tangent line at x = 0 is infinity. This means that the tangent line is a vertical line.
Therefore, the equation of the tangent line passing through (0, 0) is x = 0 or, in other words, the equation of the line is simply x = 0.
To summarize, the equation of the tangent line to the curve y = ln(x) that passes through the point (0, 0) is x = 0.