ln(o) tangent line

wferebee

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Dec 17, 2010
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I am being asked to find a tangent line of the function ln(x) at the point where x=1.
MY understanding is that to do so I can set up the equation of the line using these formulas.

1) F ' (c)= (f(b)-F(a)) / (b-a)

2) then since f ' (x) = 1/x I can set this equal to answer in step 1)

3)Then I take the point and substitute and put it in to slope intercept form. This is all fine, but the problem comes in when I plug my x coordinate (1) into the original function and get that Y=0

so....... In step 1) I obviously cant calculate ln(0) because it doesn't exist. so does that mean that my slope will be 1 or is it possible to use differentials to find an approximate value for ln(0). I used differentials seeing as a slope of one will not give me the tangent line I am looking for. Using differentials I calculated that ln(0) is close to equal to -1. Using this, my slope became 2 and the line I got out of this was pretty close to being a tangent line at x=1. My question is whether or not this is a suitable answer or is there a better way of going about finding this tangent line. I just took my final exam and have been worrying about this. Any help would be appreciated.
 
You can use y=mx+b

When x=1, then y=ln(1)=0.

\(\displaystyle y'=m=\frac{1}{x}\Rightarrow \frac{1}{1}=1\)

Sub into y=mx+b

x=1, y=0, m=1:

\(\displaystyle 0=1(1)+b\)

b=-1

The line has equation \(\displaystyle y=x-1\)
 
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