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I want to find a line that passes through $(0,-1)$ and is tangent to $\ln(x)$.

I've tried saying: ''I want to find a line that has the slope $1/x$ and passes through $(0,-1)$" but this yields:

$$y=kx+m \\ \Rightarrow 1= \frac{1}{0} +m$$

And the logic is already broken at this point.

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  • $\begingroup$ Sorry, I meant (0,-1) $\endgroup$ – user3200098 Apr 13 '14 at 20:08
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    $\begingroup$ Your flaw is plugging in $x=0$ to get the slope. The derivative gives the slope of the tangent for the value of $x$ at the point of tangency, not just anywhere. The line isn't tangent to the curve at $x=0$--in fact, $x=0$ isn't in the domain of the function being graphed. $\endgroup$ – MPW Apr 13 '14 at 20:14
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The line tangent to $\ln{x}$ at $x=a$ is

$$y=\ln{a}+\frac{1}{a}(x-a)$$

You want $y=-1$ when $x=0$, so solve for a in

$$-1=\ln{a}+\frac{1}{a}(-a)$$

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  • $\begingroup$ Why is the line given by that equation...? $\endgroup$ – user3200098 Apr 13 '14 at 20:14
  • $\begingroup$ If you are familiar with Taylor's theorem, then this is just the first two terms of the series. If not, then write it out in point slope form and solve for $y=f(x)$: $$y-y_0=m(x-x_0)$$ $$f(x)-f(a)=f'(a)(x-a)$$ $\endgroup$ – solstafir Apr 13 '14 at 20:15
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First line of your query.

Slope between $ (0,-1) , ( x, \ln x ) $ is

$ \dfrac {ln\ x +1}{x} $

which should be a maximum for tangency. So applying Quotient rule

$ \dfrac {ln\ x +1}{x} = \dfrac {1/x}{1}$

Solving, x =1, and y = 0.

Find this slope and equation of line through any of the above points.

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