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
Best Answer
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