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what are the number of tangents that can be drawn from the point $(\frac{-1}{2},0)$ to the curve $y=e^{\{x\}}$.Here { } denotes the fractional part function

what I have done:Since we cannot differentiate the fractional part function I removed the fractional part function as follows

y=$e^x$, $ x\in [0,1)$

y=$e^{x-1}$, $ x\in [1,2)$

y=$e^{x+1}$, $ x\in [-1,0)$

y=$e^{x+2}$, $ x\in [-2,-1)$

Now,just for a try I found out the tangent from the given point to curve y=$e^x$, $ x\in [0,1)$ and the equation of tangent comes out to be $y-\sqrt{e}=\sqrt{e}(x-\frac{1}{2})$.I have checked that x coordinate of point of contact of tangent on curve belongs in the interval [0,1).So,this process gives one tangent by hit and trial method but I wanted to know some general method to find the number of tangents.Please help

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  • $\begingroup$ Hint: try plotting the function, (also no need to find the equation of the tangents, you are only asked the number) $\endgroup$ – Nikunj Feb 24 '16 at 10:02
  • $\begingroup$ Hint for plotting, no need to break {x} into it's definition every time, just use that it will always lie between [0,1) for all x and hence your function lies between [1,e) for all x $\endgroup$ – Nikunj Feb 24 '16 at 10:06
  • $\begingroup$ I have rough plotted the function also but it doesn't help $\endgroup$ – Kartik Watwani Feb 24 '16 at 10:07
  • $\begingroup$ That's not possible, anyway do you have the answer to this problem? $\endgroup$ – Nikunj Feb 24 '16 at 10:09
  • $\begingroup$ yes it is 1 only $\endgroup$ – Kartik Watwani Feb 24 '16 at 10:12
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Let $x_{0}\in [n,n+1)$ and $y_{0}=e^{x_{0}-n}$, then

$y'(x_{0})=e^{x_{0}-n}$,

now the equation of tangent is

$y-e^{x_{0}-n}=e^{x_{0}-n}(x-x_{0}) \quad \cdots \cdots (*)$,

put $(-\frac{1}{2},0)$ into $(*)$,

$-e^{x_{0}-n}=e^{x_{0}-n}(-\frac{1}{2}-x_{0})$

$x_{0}=\frac{1}{2}$,

$n=0$

enter image description here

Edit: Had misread the question initially, many thanks to Tony K's comment

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  • $\begingroup$ I'm being totally intuitive here, but I think the "arcs" which are really far from the point won't have any tanget going through that point. $\endgroup$ – Soham Feb 24 '16 at 12:14
  • $\begingroup$ This is simply wrong. $\endgroup$ – TonyK Feb 24 '16 at 12:42
  • $\begingroup$ @TonyK, wait for your correct answer $\endgroup$ – Ng Chung Tak Feb 24 '16 at 12:56
  • $\begingroup$ "Therefore there's one tangent...": surely you can see that this is wrong? Whether or not I post an answer. $\endgroup$ – TonyK Feb 24 '16 at 12:59
  • $\begingroup$ @TonyK, I've rearranged the sentence structure, is that OK? $\endgroup$ – Ng Chung Tak Feb 24 '16 at 13:02
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The function is continuous and differentiable, except at integer values, where it is continuous and differentiable on the right only. At any point $x=x_0$, the (right-hand) derivative is $e^{\{x_0\}}$, and this can be taken for the slope of the tangent.

We have the tangent equation

$$y=e^{\{x_0\}}(x-x_0)+e^{\{x_0\}}=e^{\{x_0\}}(x-x_0+1).$$

It goes through the given point when

$$0=e^{x_0}(-\frac12-x_0+1)$$ or $$x_0=\frac12,\\ y=e^{1/2}(x+\frac12).$$ This is the only solution.

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