# Number of tangents from a point to a curve

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

• Hint: try plotting the function, (also no need to find the equation of the tangents, you are only asked the number) – Nikunj Feb 24 '16 at 10:02
• 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 – Nikunj Feb 24 '16 at 10:06
• I have rough plotted the function also but it doesn't help – Kartik Watwani Feb 24 '16 at 10:07
• That's not possible, anyway do you have the answer to this problem? – Nikunj Feb 24 '16 at 10:09
• yes it is 1 only – Kartik Watwani Feb 24 '16 at 10:12

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 $(*)$,

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

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

$n=0$

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

• 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. – Soham Feb 24 '16 at 12:14
• This is simply wrong. – TonyK Feb 24 '16 at 12:42
• @TonyK, wait for your correct answer – Ng Chung Tak Feb 24 '16 at 12:56
• "Therefore there's one tangent...": surely you can see that this is wrong? Whether or not I post an answer. – TonyK Feb 24 '16 at 12:59
• @TonyK, I've rearranged the sentence structure, is that OK? – Ng Chung Tak Feb 24 '16 at 13:02

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.